Calc I MasterProblemSheet
Calc I MasterProblemSheet
Calc I MasterProblemSheet
Calculus I
This master problem sheet contains all freecalc problems on the topics studied in Calculus I. For a list of contributors/authors of
the freecalc project (and in particular, the present problem collection) see the following file. https://sourceforge.net/p/
freecalculus/code/HEAD/tree/trunk/contributors.tex
These lecture slides and their LATEX source code are licensed to you under the Creative Commons license CC BY 3.0. You are free
Contents
1 Functions, Basic Facts 2
1.1 Understanding function notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Domains and ranges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Piecewise Defined Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 Function composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.5 Linear Transformations and Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Trigonometry 9
2.1 Angle conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Trigonometry identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Trigonometry equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4 Inverse Functions 17
4.1 Problems Using Rational Functions Only . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.2 Problems Involving Exponents, Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1
6 Derivatives 23
6.1 Derivatives and Function Graphs: basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.2 Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
6.3 Basic Trigonometric Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
6.4 Natural Exponent Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
6.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
6.6 Problem Collection All Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
6.7 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
6.8 Implicit Differentiation and Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
6.9 Derivative of non-Constant Exponent with non-Constant Base . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6.10 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
9 Integration Basics 50
9.1 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
9.2 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
9.3 Basic Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
9.4 Fundamental Theorem of Calculus Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
9.5 Integration with The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
9.5.1 Substitution in Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
9.5.2 Substitution in Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
f (2 + h) − f (2) f (a + h) − f (a)
1. , where f (x) = x2 − x − 1. 4. , where f (x) = x4 .
h h
answer: h + 3 answer: 6a2 h + 4ah2 + h3 + 4a3
2
x+4
1. f (x) = . answer: x ∈ [−1, 5].
x2 − 4
x ∈ (−∞, −2) ∪ (−2, 2) ∪ (2, ∞)
1
alternatively: answer:
5. h(x) = √
6
.
x2− 7x
x 6= ±2,
2x3 − 5
answer: x ∈ (−∞, 0) ∪ (7, ∞).
2. f (x) = .
x2 + 5x + 6 u+1
x ∈ (−∞, −3) ∪ (−3, −2) ∪ (−2, ∞) 6. f (u) = 1 .
alternatively:
x 6= −2, −3,
answer: 1 + u+1
√
u ∈ (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞)..
answer:
u 6= −1, −2 or
3
3. f (t) = 3t − 1.
√
q
answer: x ∈ R (the domain is all real numbers)
7. F (x) = 10 − x.
√ √
4. g(t) = 5−t− 1 + t. answer: x ∈ [0, 100]
1 x
1
1 x
1.
3.
3 3
if 2 < x ≤ 6 4x − 8
2 answer: y = if 2 ≤ x < 3 5x − 12
if 0 < x ≤ 2 −3x + 3 4 2 answer: y =
( if − 2 ≤ x < 2 −3x − 1
(
y y
1 1
x 1 x
1
2. 4.
if 1 ≤ x ≤ 2 3x − 4
if 0 ≤ x < 1 −2x + 1
if 1 < x ≤ 2 −4x + 6 answer: y =
if − 1 ≤ x < 0 2x + 1
4 4 answer: y =
if − 3 ≤ x < 1 3x + 5 if − 2 ≤ x < −1 −3x − 4
(
Problem 4.
Write down formulas for function whose graphs are as follows. The graphs are up to scale. All arcs are parts of circles.
y
2
(−4, 1) (4, 1)
1
x
1. −4 −3 −2 −1
−1 1 2 3 4
Problem 5. Plot the piecewise defined functions by hand. Compare your answer to the plot of a computer algebra system.
1. G(x) = x+|x|
2x . 3. f (x) =
x x≤1
.
x2 x ≥ 1
2. g(x) = |x| − x.
3
1. f (x) = x2 + 1, g(x) = x + 1. in some order: (1 + x)2 + 1, (x)2 + 2, ((x)2 + 1)2 + 1, 2 + x
Domain, all 4 cases: x ∈ R (all reals)
answer:
√ in some order: 2 + x, 1 + 1 + x, 1 + 1 + x, 2 + x
2. f (x) = x + 1, g(x) = x + 1. Domain of f ◦ f is x ≥ −1. Domain of p
√ √ √
g ◦ g is all reals (x ∈ R). answer:
Domain of f ◦ g is x ≥ −2. Domain of g ◦ f is x ≥ −1
x+1 x−1 x x
in some order: −x, 1 , x, − 1
4. f (x) = , g(x) = . answer:
x−1 x+1
Domain f ◦ g: x 6= −1. Domain g ◦ f : x 6= 0, x 6= 1
Domain f ◦ f : x 6= 1. Domain g ◦ g: x 6= 0, x 6= −1
Problem 7. Compute the composite functions (f ◦ g)(x), (g ◦ f )(x). Simplify your answer to a single fraction. Find the domain of the
composite function. The answer key has not been fully proofread, use with caution.
x+2 x−1
1. f (x) = , g(x) = .
x−2 x+2
3 −2+3x
x 6= 2, 2 (g ◦ f )(x) =
4
−5−x answer:
x 6= −2, −5 (f ◦ g)(x) = 3+3x
x+1 x−2
2. f (x) = , g(x) = .
3x − 2 x−1
3 2 3−2x
x 6= 2 , 3 (g ◦ f )(x) = 5−5x
−4+x answer:
x 6= 4, 1 (f ◦ g)(x) = −3+2x
2x + 1 x−2
3. f (x) = , g(x) = .
3x − 1 2x − 1
3 3+x
x 6= −3, 1 (g ◦ f )(x) = 3−4x
2 −5+x answer:
x 6= 5, 1 (f ◦ g)(x) = −5+4x
x+1 x+2
4. f (x) = , g(x) = .
x−2 2x − 1
4+x
x 6= −4, 2 (g ◦ f )(x) = −3+3x
3 2 4−3x answer:
x 6= 4 , 1 (f ◦ g)(x) = 1+3x
5x + 1 4x − 1
5. f (x) = , g(x) = .
4x − 1 3x + 1
19 4 2+19x
x 6= − 2 , 1 (g ◦ f )(x) = 5+16x
13 3 −5+13x answer:
x 6= 5 , − 1 (f ◦ g)(x) = −4+23x
3x − 5 x−2
6. f (x) = , g(x) = .
x−2 x−4
−x+3
x 6= 3, 2 (g ◦ f )(x) = x−1
−x+6 answer:
x 6= 6, 4 (f ◦ g)(x) = −2x+14
x−3 y+3
7. f (x) = , g(y) = .
x+2 y−4
3 −3x−11
x 6= − 11 , −2 (g ◦ f )(x) =
4x+3
3 3x−5 answer:
x 6= 5 , 4 (f ◦ g)(x) = −2x+15
1 3
1. y = . 4. y = .
x 2x + 1
1 3+x
2. y = . 5. y = .
x+1 2x + 1
1 3+x
3. y = . 6. y =
.
2x + 1 2x + 1
Problem 9. Sketch by hand approximately the given function. The function is obtained by transforming linearly the graph of a known
function. The known function has been sketched for you by computer.
4
1
1. f (x) = − x2 + 1.
2
y = x2
answer:
1 2
2. f (x) = x + x − 1.
2
y = x2
answer:
5
1
3. f (x) = + 1.
2x − 1
1
y= x
answer:
1
2x+ 14 1
4. f (x) = + .
x − 12 2
1
y= x
answer:
6
√
5. f (x) = − 2x − 1 − 1
√
y= x
answer:
y =
√
x
√
6. f (x) = − −2x − 1 + 1
√
y= x
answer:
y =
√
x
Solution. 9.3.
1
We are asked to plot f (x) = + 1 by linearly transforming the graph of g(x) = x1 to the graph of f (x). To do that we have to
2x − 1
compose g with a sequence of linear transformations to obtain f (x). There are two natural ways to do that; we show both by presenting
two different solutions.
Solution I. We show how to get from g(x) = x1 to f (x) by composing g with a sequence of linear transformations.
1
g(x) = x
1
Define h(x) via: h(x) = g(x + 1) =
x−1
1
Define k(x) via: k(x) = h(2x) =
2x − 1
1
Therefore f (x) = k(x) + 1 = +1
2x − 1
7
We plot consecutively the functions g(x), h(x), k(x) and f (x). We start from the given graph of g(x).
1
y = h(x) = g(x − 1) = x−1
1 1
y = k(x) = h(2x) = 2x−1 y = f (x) = k(x) + 1 = 2x−1 +1
Solution II. In the previous solution we used horizontal stretch to transform the graph of h(x) to the graph of k(x) = h(2x). Algebra
suggests a second way to transform the graph of g(x) to the graph of f (x), this time using a vertical stretch. Indeed, we have the
equality
1 1 1
f (x) = +1= · + 1.
2x − 1 2 x − 21
Therefore we can carry out the sequence of transformations shown below.
1
g(x) = x
1 1
Define l(x) via: l(x) = g x− =
2 x − 21
1 1 1 1
Define k(x) via: k(x) = h(x) = · 1
=
2 2 x− 2 (2x − 1)
1
Therefore f (x) = k(x) + 1 = +1
2x − 1
8
We plot consecutively the functions g(x), l(x), k(x) and f (x). We start from the given graph of g(x).
1 1
y = l(x) = g x − 2 = x− 12
y = k(x) = 21 l(x) = 1
2x−1 y = f (x) = k(x) + 1 = 1
2x−1 +1
2 Trigonometry
2.1 Angle conversion
Problem 10. Convert from degrees to radians.
2. 30◦ .
9. 135◦ . 15. 360◦ .
6
answer: π ≈ 0.523598776
4
answer: 3π answer: 2π
3. 36◦ .
5
answer: π ≈ 0.628318531 10. 150◦ . 16. 405◦ .
4. 45◦ .
6 4
answer: 5π answer: 9π
◦
5. 60 . answer: π
3
answer: 20π
3
answer: π ≈ 1.047197551
Problem 11. Convert from radians to degrees. The answer key has not been proofread, use with caution.
7
1. 4π. 3. 12 π. 5. − 38 π.
answer: 720◦ answer: 105◦ answer: −67.5◦
2. − 76 π. 4. 4
3 π.
6. 2014π.
answer: −210◦ answer: 240◦ answer: 362520◦
9
7. 5. 8. −2014.
π
900 ◦ ≈ 286◦ answer: answer: −362520◦
√
2. sin(2x) = cos x. 2
√ 2
7. 2 cos x − (1 + 2) cos x + = 0.
2 2 6 6 2
answer: x = π , x = 3π , x = π , or x = 5π 4 3 3 4
answer: x = π , π , 5π , 7π .
√
3. 3 sin x = sin(2x). 6 6
answer: x = π , 11π , 0, π, 2π . 8. | tan x| = 1.
4 4 4 4
answer: x = π , x = 3π , x = 5π , or x = 7π
4. 2 sin2 x = 1.
4 4 4 4
9. 3 cot2 x = 1.
answer: x = π , x = 3π , x = 5π , or x = 7π
3 3 3 3
answer: x = π , x = 2π , x = 4π , or x = 5π
10
√
1 2 π 5π
Therefore u = cos x = 2 or u = cos x = 2 , and, as x is in the interval [0, 2π], we get x = 3, 3 (for cos x = 12 ) or x = π 7π
4, 4 (for
√
cos x = 22 ).
√
x2 − 3x x2 + 16 − 5
2. lim 2
. 13. lim .
x→3 x − 2x − 3 x→−3 x+3
4 5
answer: 3 answer: − 3
x2 − 5x − 6 1
+ x1
3. lim . 14. lim 3
.
x→2 x−2 x→−3 3+x
answer: DNE 9
answer: − 1
x2 − 3x x2 + 4x + 4
4. lim . 15. lim .
x→−1 x2 − 2x − 3 x→−2 x4 − 16
answer: DNE
answer: 0
√ √
x2 − 4 1+x− 1−x
5. lim . 16. lim .
x→−2 2x2 + 5x + 2 x→0 x
3
answer: 4 answer: 1
2x2 + 3x + 1 1 1
6. lim . 17. lim − .
x→−1 3x2 − 2x − 5 x→0 x x2 + x
8 answer: 1
answer: 1
√
(−3 + h)2 − 9 3− x
7. lim . 18. lim .
x→9 9x − x2
h→0 h 54
answer: 1
answer: −6
x+3 1 1
9. lim . 20. lim √− .
x→−3 x3 + 27 x→0 x 1+x x
27 2
answer: 1 answer: − 1
x4 − 1 (x + h)3 − x3
10. lim . 21. lim .
x→1 x3 − 1 h→0 h
3 answer: 3x2
answer: 4
√ 1 1
4+h−2 (x+h)2 − x2
11. lim . 22. lim .
h→0 h h→0 h
4 x
answer: 1 answer: − 23
Solution. 14.22
1 1
(x+h)2 − x2 x2 − (x + h)2 x2 − (x2 + 2xh + h2 )
lim = lim 2 2
= lim
h→0 h h→0 hx (x + h) h→0 hx2 (x + h)2
h
(−2x + h) −2x + 0 2
= lim 2 (x + h)2
= 2 2
= − 3.
h→0 h
x x (x + 0) x
11
3x2 + 4x − 7
1. lim answer: 5.
x→1 x3 − x
2x2 − 3x − 5
2. lim 3
answer: − 7 .
x→−1 x3 + 1
Problem 16. Evaluate the limits. Justify your computations.
1 √
1. lim 2x2 − 3x − 6. answer: −4
3. lim . 6 5. lim (1 + 3
x)(2 − x). answer: −18
x→2 x→−1 x2 − 3x + 2
answer: 1
x→8
x4 − x p
2. lim . answer: 1 4. lim x4 + 16. 32 answer:
x→−1 x2 + 2x + 3 x→−2 √
1 − x2 √ p
2. lim . 16x6 − 3x 18. lim x2 + 1.
x→∞ x3 − x − 1 10. lim . x→∞
answer: 0
x→−∞ x3 + 2
answer: ∞
answer: −4
x−2
3. lim . p 19. lim (x4 + x5 ).
x→−∞ x2 + 5 11. lim 4x2 + x − 2x. x→−∞
answer: 0
x→∞
4 answer: −∞
answer: 1
3x3 + 2 √
4. lim . p 1 + x6
x→−∞ 2x3 − 4x + 5 12. lim x + x2 + 3x. 20. lim .
2
answer: 3 x→−∞ x→−∞ 1 + x2
2
√ answer: − 3 answer: ∞
x + x2 √ √ √
5. lim √ . 21. lim (x − x).
x→∞ x − x2 13. lim x2 + 2x − x2 − 2x. x→∞
x→∞
answer: −1
√
answer: 2. answer: ∞
3−x x √ √
6. lim 3 . 14. lim x2 + x − x2 − x. 22. lim (x2 − x3 ).
x→∞ 2x 2 − 2 2
x→−∞ x→∞
answer: − 1
answer: −1. answer: −∞
(2x2 + 3)2 p p
7. lim . 15. lim x2 + ax − x2 + bx. 23. lim x sin x.
x→∞ (x − 1)2 (x2 + 1) x→∞ x→∞
answer: 4 2 answer: DNE
answer: a−b
x2 − 3 √
8. lim √ . 16. lim cos x. 24. lim x sin x.
x→∞ x→∞
x→∞ x4 + 3
answer: 1 answer: DNE answer: DNE
Solution. 17.4.
3x3 + 2 x13
3x3 + 2 Divide top
lim = lim
x→−∞ 2x3 − 4x + 5 x→−∞ (2x3 − 4x + 5) 13 and bottom
x
3 + x23 by highest term
= lim
x→−∞ 2 − 42 + 53 in denominator
x x
3+0 3
= = .
2−0+0 2
12
Solution. 17.14.
p p p p √x 2 + x + √ x 2 − x
lim x2 + x − x2 − x = lim x +x− x −x √
2 2 √
x→−∞ x→−∞ x2 + x + x2 − x
x2 + x − (x2 − x) 2x x1
= lim √ √ = lim √ √
x2 + x + x2 − x x1
x→−∞ x2 + x + x2 − x x→−∞
2 2
= lim √ 2 √ = lim q q
x→−∞ x +x x2 −x x→−∞ 2 x2 −x
x + x − x x+x2 − x2
2 2
= lim q q = √ √ = −1
x→−∞
− 1+ 1 − 1− 1 − 1+0− 1−0
x x
√
The sign highlighted in red arises from the fact that, for negative x, we have that x = − x2 .
q
x2
4 −1
2. lim . answer: ∞.
x→−2− 2x2 + 3x − 2
(+)
where the latest term is +∞ because it is of the form (+)(+) .
2x
1. y = √ . answer: vertical: x = 2, horizontal: y = 3
x2 + x + 3 − 3
answer: Vertical: x = 2, x = −3, horizontal: y = 2, y = −2
x2 − 1
4. y = .
2x2 − 3x − 2
3x2
2. y = √ . 2 2
answer: vertical: x = 2, x = − 1 , horizontal: y = 1
x2 + 2x + 10 − 5
2x2 − 2x − 1
answer: Vertical: x = 3, x = −5, horizontal: none.
5. y = .
3x + 1 x2 + x − 2
3. y = .
x−2 answer: vertical: x = 1, x = −2, horizontal: y = 2
13
1 + x4 x−9
6. y = . 8. y = √ .
x2 − x4 4x2+ 3x + 3
2
answer: vertical: x = 0, x = 1, x = −1, horizontal: y = −1 answer: no vertical asymptote, horizontal: y = ± 1
√
x3 − x x2 + 1 − x
7. y = 2
. 9. y = .
x − 7x + 6 x
answer: vertical: x = 6, no horizontal asymptote answer: vertical: x = 0, horizontal: y = 0, y = −2,
Solution. 20.1 Vertical asymptotes. A function f (x) has a vertical asymptote at x = a if lim f (x) = ±∞.
x→a
The function is algebraic, and therefore, if it is defined, has a finite limit (i.e., no asymptote).√Therefore the function can have
vertical asymptotes only for those x for which f (x) is not defined. The function is not defined for x2 + x + 3 − 3 = 0, which has
two solutions, x = 2 and x = −3. These are precisely the vertical asymptotes: indeed,
2x 2x
lim+ √ =∞ lim− √ = −∞
x→2 x2 +x+3−3 x→2 x2 +x+3−3
and
2x 2x
lim √ =∞ lim − √ = −∞
x→−3+ +x+3−3 x2 x→−3 2
x +x+3−3
Horizontal asymptotes. A function f (x) has a horizontal asymptote if lim f (x) exists. If that limit exists, and is some number,
x→±∞
say, N , then y = N is the equation of the corresponding asymptote.
Consider the limit x → −∞. We have that
2x 2
lim √ = lim √ 2
2
x + 3x + 3 − 3 x +x+3
x→−∞ x→−∞
x − x3
2 1
q
= lim q x = − x12 when x < 0
x→−∞ x2 +3x+3
− x2 − x3
2
= lim q
x→−∞
− 1 + x3 + x32 − x3
lim 2
x→−∞
= q
− lim 1 + lim x3 + lim x32 − lim x3
x→−∞ x→−∞ x→−∞ x→−∞
2
= √
− 1+0+0−0
= −2 .
Therefore y = −2 is a horizontal asymptote.
The case x → ∞, is handled similarly and yields that y = 2 is a horizontal asymptote.
A computer generated graph confirms our computations.
x = −3
y = q 2x
x2 +x+3−3
y = 2
10
y = −2
x = 2
14
x2 − 4 2(x + h)3 − 2x3
1. lim 3
answer: 4 4. lim answer: 6x2
x→2 x2 −x−2 h→0 h
√
5x3 + x − 1 9x2 − 2
2. lim 2
answer: 5 5. lim
2x3 − 7
answer: 3
x→−∞ x→∞ x+4
x−3 2x + 3
3. lim+ answer: −∞ 6. lim answer: Does not exist
x→1 x−1 x→−1 x+1
3.4 Continuity
3.4.1 Continuity to evaluate limits
Problem 22. Use continuity to evaluate the limits. The answer key has not been proofread, use with caution.
• limπ x tan x.
x→ 4
4
answer: π
1
• lim √
x→0 1− 3 + cos x
answer: −1
Problem 24. Show that f (x) is continuous at all irrational points and continuous at all rational ones.
if x is rational and x = pq
1
f (x) = q2
0 if x is irrational
where in the first item p, q are relatively prime integers (i.e., integers without a common divisor).
x2 − x − 6 x3 − 27
1. f (x) = . 2. f (x) = .
x−3 x2 − 9
with domain x ∈ (−∞, −3) ∪ (−3, ∞).
x+3
f¯(x) = x +3x+9
2 answer:
Extend f (x) to
Extend f (x) to f¯(x) = x + 2. x ∈ (−∞, −3) ∪ (−3, 3) ∪ (3, ∞).
answer:
Implied domain: x ∈ (−∞, 3) ∪ (3, ∞). Implied domain:
Problem 26. Find the numbers x for which f is discontinuous. At which of these numbers is f continuous from the right, from the left,
or neither?
2 + x2 if x ≤ 0 x + 2 if x < 0
1. f (x) = −2x if 0 < x ≤ 2 . 3. f (x) = 2x2 if 0 ≤ x ≤ 1 .
2
−x if x > 2 2 − x if x > 1
x+1 if x ≤ 1
1
2. f (x) = if 1<x<2 .
√
x
x − 2 if x ≥ 2
15
Problem 27. Find the values of a and b that make f continuous everywhere.
1 if x < 0
1. f (x) = ax + b if 0 ≤ x < 1 .
2x if x ≥ 1
x2 − 1
if x < 1
2. f (x) = x−1 .
2
ax − bx + 3 if 1 ≤ x < 3
2x − a + b if x ≥ 3
Problem 29.
1. (a) Solve the equation x2 + 13x + 41 = 1.
(b) Use the intermediate value theorem to prove that the equation x2 + 13x + 41 = sin x has at least two solutions, lying
between the two solutions to 29.1.a.
where for the very last inequality we use the fact that sin 6.5 < 1 (remember sin t ≤ 1 for all real values of t).
On the other hand,
f (−5) = g(−5) − sin(−5) = 1 + sin 5 > 0
as sin 5 > −1 (remember sin t ≥ −1 for all real values of t). Therefore f (−5) > 0 and f (−6.5) < 0 and by the Intermediate Value
Theorem (IVT) f (x) = 0 has a solution in the interval x ∈ (−6.5, −5).
Proving f (x) = 0 has a solution in the interval x ∈ (−8, −6.5) is similar and we leave it to the student.
16
Below is a computer generated plot of the function with the use of which we can visually verify our answer.
y
y = x2 + 41 + 13x − sin x
x
−8 −6.5 −5
y = x2 + 13x + 40
4 Inverse Functions
4.1 Problems Using Rational Functions Only
Problem 30. Find the inverse function. You are asked to do the algebra only; you are not asked to determine the domain or range of
the function or its inverse.
1. f (x) = 3x2 + 4x − 7, where x ≥ − 23 .
3 3 3
x ≥ − 25 , answer: f −1 (x) = − 2 +
25+3x
√
2x + 5
3. f (x) = , where x 6= 4.
x−4
x−2
x 6= 2 answer: f −1 (x) = 4x+5 ,
3x + 5
4. f (x) = , where x 6= 2.
2x − 4
2 2x − 3
x 6= , (x) = answer: f
3 4x + 5 −1
5x + 6
5. f (x) = , where x 6= − 54 .
4x + 5
4x−5 4
answer: f −1 (x) = −5x+6 , x 6= 5
2x − 3
6. f (x) = , where x 6= 43 ..
−3x + 4
3x+2 3
answer: f −1 (x) = 4x+3 , x 6= − 2
17
Solution. 30.4 This is a concise solution written in form suitable for test taking.
3x + 5
y =
2x − 4
y(2x − 4) = 3x + 5
2xy − 4y = 3x + 5
2xy − 3x = 4y + 5
x(2y − 3) = 4y + 5
4y + 5
x =
2y − 3
5 + 4y
Therefore f −1 (y) =
2y − 3
5 + 4x
f −1 (x) = .
2x − 3
as expected.
Problem 31. Find the inverse function f −1 . Plot roughly by hand y = f (x). Using the plot of y = f (x), plot roughly by hand f −1 (x).
Indicate the relationship between the graph of f (x) and f −1 (x).
1. f (x) = x2 + 2x − 2, x ≥ −1. x+3−1
√
answer: f −1 (x) =
2. f (x) = x2 + x − 2, x ≥ − 21 .
2
answer: f −1 (x) =
4x+9−1
√
answer: f −1 (x) = ex − 3
3. y = (ln x)2 , x ≥ 1.
3
2. f (x) = ex . x ≥ 0 answer: f −1 (x) = e x ,
√
18
ex 5. f (x) = 22x + 2x − 2.
4. y = .
1 + 2ex
2 1−2x 2
0, 1 x ∈ , answer: f −1 (x) = ln x ≥ −2 answer: f −1 (x) = log2 ,
x −1+ 9+4x
√
Solution. 32.1
y = ln(x + 3)
ey = eln(x+3)
ey = x + 3
ey − 3 = x
Therefore f −1 (y) = ey − 3.
The domain of ey is all real numbers, so the domain of f −1 is all real numbers.
Solution. 32.3 √
y
= (ln x)2 take on both sides, y ≥ 0
√
√
y
= ln x exponentiate
e y = eln
√
x
=x
f −1 (y) = e√ y
f −1 (x) = e x
Solution. 32.4
ex
y=
1 + 2ex
y(1 + 2e ) = ex
x
y = ex (1 − 2y)
y
= ex
1 − 2y
y
ln = ln ex
1 − 2y
y
ln =x
1 − 2y
y
Therefore f −1 (y) = ln .
1 − 2y
The natural logarithm function is only defined for positive input values. Therefore the domain is the set of all y for which
y
> 0.
1 − 2y
This inequality holds if the numerator and denominator are both positive or both negative. This happens if either
1. y > 0 and y < 12 , or
2. y < 0 and y > 12 .
The latter option is impossible, so the domain is {y ∈ R | 0 < y < 12 }.
25 · 27
1. √
2 2
answer: 210.5 = 2 2
21
32 · 3−1
2. √
33 · 33
answer: 3 2
−7
19
π3
3. √
π −1 π 5
answer: π 2
3
Solution. 33.2.
32 · 3−1 32 · 3−1
√ = 1
33 · 33 33 · (33 ) 2
32 · 3−1
= 3
33 · 3 2
32−1
= 3+ 3
3 2
31
= 9
32
9
= 31− 2
7
= 3− 2 .
1. log2 16. 3
answer: − 4
answer: 4 √
5. log2 (8 2).
1
2. log3 .
2
answer: 7
9
answer: −2
6. log 12 (4).
3. log10 1000. answer: −2
√
7. log 19 ( 3).
answer: 3
2
4. log6 36− 3 . 4
answer: − 1
Solution. 36.13.
49x
log7 343y = log7 49x − log7 343y
= x log7 49 − y log7 343
49x
However 49 = 72 and 343 = 73 , therefore log7 343y = 2x − 3y.
Problem 35. Express each of the following as a single logarithm. If possible, compute the logarithm without using a calculator. The
answer key has not been proofread, use with caution.
1. ln 4 + ln 6 − ln 5.
5
24 answer: ln
2. 2 ln 2 − 3 ln 3 + 4 ln 4.
27
1024 answer: ln
3. ln 36 − 2 ln 3 − 3 ln 2.
2
1 answer: − ln 2 = ln
20
5. log7 (24) + log 17 3 − log49 (64).
answer: 0
3
6. log3 (24) + log3 8 .
answer: 2
Solution. 35.2.
2 ln 2 − 3 ln 3 + 4 ln 4 = ln 22 − ln 33 + ln 44
= ln 4 − ln 27 + ln 256
4
= ln + ln 256
27
4 · 256
= ln
27
1024
= ln .
27
1024
is not a rational power of e, therefore ln 1024
27 27 is not a rational number and there are no further simplifications of the answer
(except possibly a numerical approximation with a calculator or equivalent).
Solution. 35.5
=0.
1
2. log3 . 7. log1.5 2.25.
27
answer: −3 answer: 2
1 8. log5 4 − log5 500.
3. ln .
e
answer: −1 answer: −3
√
4. log10 10. 9. log2 6 − log2 15 + log2 20.
2
answer: answer: 3
1
21
11. e−2 ln 5 . answer: 10
49x
25
answer: 1
13. log7
10
343y
12. ln ln ee . answer: 2x − 3y
Solution. 36.13.
49x
log7 343y = log7 49x − log7 343y
= x log7 49 − y log7 343
49x
However 49 = 72 and 343 = 73 , therefore log7 343y = 2x − 3y.
2. ln(2x − 9) = 2. x
11. ee = 10.
2
answer: e +9 ≈ 8.194528
2
answer: ln(ln 10) ≈ 0.834
3. ln(x2 − 2) = 3.
12. ln(2x + 1) = 3 − ln x.
answer: ± e3 + 2 ≈ ±4.699525
q 4
≈ 2.928878 answer:
−1+ 1+8e3
4. 2x−3 = 5. q
ln 2
answer: log2 5 + 3 = ln 5 + 3 ≈ 5.321928 13. e2x − 4ex + 3 = 0.
6. e2x+1 = t. answer: x = 0
2
15. e2x − ex − 6 = 0.
answer: ln t−1
−2x
8. e − e = 1. 3
answer: x =
log2 5
2
answer: − 1 ln(e − 1) ≈ −0.271
1 x−1
9. 8(1 + e−x )−1 = 3. 17. 3 · 2x + 2 2 − 7 = 0.
3 5 ln 2
answer: − ln 5 = ln 3 ≈ −0.510826 answer: x = 0 or 2 − log2 3 = 2 − ln 3
Solution. 37.4
2x−3 = 5 take log2
x−3 = log2 (5) add 3 to both sides
x = log2 (5) + 3 answer is complete
ln 5
= +3 optional step: convert to ln
ln 2
≈ 5.321928095 calculator
Solution. 37.8
22
e − e−2x = 1
e−2x = e−1 apply ln
ln e−2x = ln(e − 1)
−2x = ln(e − 1)
1
x = − ln(e − 1)
2
≈ −0.270662427 calculator
Solution. 37.5
ln x + ln(x − 1) = 1
ln(x2 − x) = 1
2
−x)
eln(x = e1
x2 − x = e
x2 − x − e = 0
p
−(−1) ± (−1)2 − 4(1)(−e)
Quadratic formula: x=
2(1)
√
1± 1 + 4e
= .
2
√ √ √
1− 1+4e 1− 1+4e 1+ 1+4e
However 2 is negative, so ln 2 is undefined. Hence the only solution is x = 2 ≈ 2.2229.
6 Derivatives
6.1 Derivatives and Function Graphs: basics
Problem 38. Match the graph of each the following functions
1. 3.
2. 4.
23
1. 3.
2. 4.
Give reasons for your choices. Can you guess formulas that would give a similar (or precisely the same) graph, and confirm visually
your guess using a graphing device?
answer: 0
7. g(x) = x2 (1 − 2x).
2015
2. f (x) = π .
answer: 2x − 6x2
answer: 0
3 9. f (x) = 2x− 4 .
4. f (x) = x8 .
4 4
−7
2
answer: − 3 x
answer: 6x7
1 6 12
6. f (t) = t − 3t4 + t. 11. A(x) = − . answer: 60x−6
2 x5
Solution. 39.7 Approach
0 1. Uncover
0 the parenthesis, and then differentiate:
x2 (1 − 2x) = x2 − 2x3 = 2x − 6x2
Approach 2. Use
0 first the2product rule and then simplify:
x2 (1 − 2x) = (x )0 (1 − 2x) + x2 (1 − 2x)0
= 2x(1 − 2x) + x2 (−2)
= 2x − 4x2 − 2x2
= 2x − 6x2 .
Of course, both approaches lead to the same answer.
24
Problem 40. Compute the derivative.
5 2 √ √
1. y = x 3 − x 3 . 10. y = 5
x + 4 x5 .
3 5
answer: 5 x 3 − 2/3x 3 answer: 10x 2 + 1 x 5
2 −1 3 −4
√ 2
x − x. √
2. f (x) = 1
11. y = x+ √ .
2
answer: −1 + 1 x 2
3
x
−1 3 3
answer: 1 + 1 x 6 − 2 x 3
−5 −5
√
3. y = x(x − 1).
12. f (x) = (1 + 2x2 )(x − x2 ).
2 2
answer: 3 x 2 − 1 x 2
1 −1
answer: 1 − 2x + 6x2 − 8x3
5. f (x) = 4πx2 .
−5
x2 + 4x + 3
answer: −6 − 4x3 + 14x6
6. y = √ .
x 15. f (x) = (1 + x + x2 )(2 − x4 ).
2 2
answer: 2x 2 + 3 x 2 − 3 x 2
−1 1 −3
answer: 2 + 4x − 4x3 − 5x4 − 6x5
√
x+x
7. y = . 1 3
x2 16. g(y) = − 4 (y + 5y 3 ).
2
answer: −x−2 − 3 x 2
y2 y
−5
answer: 5 + 9y −4 + 14y −2
3
8. f (x) = x + x−1 . 17. f (x) = (x3 − 2x)(x−4 + x−2 ).
answer: 3x2 + 3 − 3x−2 − 3x−4 answer: 1 + 6x−4 + x−2
√ √
9. f (x) = 2x + 5x. 1 + 2x
18. f (x) = .
2 2 x
3 − 4x
5 x− 2 = √2 + √5 2+
√
answer:
answer: 10(3 − 4x)−2
1 √ √
Solution. 40.11
2 !0 2 0
√
1 1 1
x+ √ = x 2 + x− 3
3
x
1 1 1
2 1 2 0
= + 2x 2 x− 3 + x− 3
x2
1 2
0
= x + 2x 6 + x− 3
1 1 2 2
= 1 + 2 · x 6 −1 + − x− 3 −1
6 3
1 5 2 5
= 1 + x− 6 − x− 3 .
3 3
x3 t
2. y = . 5. y = .
1 − x2 (t − 1)2
(1−x2 )2 (t−1)3
answer: 3x −x answer: −
2 4 t+1
x+1 t2 + 2
3. y = . 6. y = .
x3 +x−2 t4 − 3t2 + 1
(−2+x+x3 )2 (1−3t2 +t4 )2
answer: −3−3x −2x answer: 14t−8t
2 3 3 −2t5
25
√
t− t u6 − 2u3 + 5
7. g(t) = . 13. y = .
t
1
3 u2
6 3
answer: − 1 t 6 + 2 t 3 answer: −2 − 10u−3 + 4u3
−5 −1
8. y = ax2 + bx + c. ax + b
14. f (x) = .
answer: b + 2ax cx + d
(d+cx)
answer: ad−bc2
B C
9. y = A + + 2.
x x x3
answer: −Bx−2C 1+x
15. f (x) = .
1 + x2
2t
10. f (t) = √ . (2+x)2
2+ t
answer: x +4x+2
2
2+t 2
1 2
! answer: 1+x
16. f (x) =
4+t 2
1 .
1 + x3
cx
11. y = . (3+x)2
1 + cx
answer: x +6x+3
2
answer: c(1 + cx)−2
√ x
12. y =
3
t(t2 + t + t−1 ). 17. f (x) = c .
x+ x
3 3 3 (c+x2 )2
answer: − 2 t 3 + 4 t 3 + 7 t 3
−5 1 4 2xc answer:
Solution. 41.5 This can be differentiated more effectively using the chain rule, however let us show how the problem can be solved
directly using the quotient rule. 0
0
(t)0 (t − 1)2 − t (t − 1)2
t
=
(t − 1)2 (t − 1)4 0
(t − 1) − t t2 − 2t + 1
2
=
(t − 1)4
(t − 1)2 − t (2t − 2)
=
(t − 1)4
−
(t
1) ((t − 1) − 2t)
=
(t − 1)4 3
−t − 1
=
(t − 1)3
t+1
= −
(t − 1)3
√
2. f (x) = x cos x. 7. y = c cos t + t2 sin t.
2 answer: −c sin t + 2t sin t + t2 cos t
answer: −x 2 sin x + 1 x 2 cos x
1 −1
26
(1+cos θ)2 1 − sec x
sin θ answer:
14. y = .
tan x
sin2 x
cos x answer: cos x−1
12. y = .
1 − sin x
1−sin x
15. h(θ) = θ csc θ − cot θ.
1 answer:
sin2 θ
answer: 1+sin θ−θ cos θ
t sin t
13. y = . 16. y = x2 sin x tan x.
1+t
(1+t)2 cos2 x
answer: sin t+t cos t+t answer:
2 cos t 2x cos x sin2 x+2x2 sin x cos2 x+x2 sin3 (x)
1. tan x. 8. csc2 x.
answer: sec2 x
answer: −2 cot x csc2 x
2. cot x.
sin x
answer: − csc2 x
9. .
x
3. sec x. x2
answer: x cos x−sin x
cos2 x
sin x answer: sec x tan x =
4. csc x. sin x
10. .
sin x
ex
answer: − csc x cot x = − cos2x
ex
answer: cos x−sin x
5. sec x tan x.
answer: sec x tan2 x + sec3 x 11. x sin xex .
Solution.
2
r = Ae−kt .
Let u = −kt2 .
Then r = Aeu .
dr dr du
Chain Rule =
dt du dt
= (Aeu )(−2kt)
2
= −2Akte−kt .
ex
3. y = (sin x + cos x).
2
27
p
1. f (x) = 1 + x2 10. y = cos (4x)
answer: x(x2 + 1) 2 answer: −4 sin (4x)
−1
1
3. y = (cos x) 2
answer: −12
sin 3x2
3
x cos 3x2
2 x 2
x = √ answer: 1 x 2 cos
√ cos x −1
√
x
13. 23 .
4. f (x) = sin3 x.
answer: 23 3x (ln 2)(ln 3)
answer: 3 cos x sin2 x x
1
6. f (x) = . p
sin3 x sin4 x
15. y = sec(4x)
answer: − 3 cos x
√
answer: 2 sec(4x) tan(4x) = 2(sec(4x)) 2 sin(4x).
p 3
3
7. f (x) = 4 + 3 tan x.
answer: (4 + 3 tan x) 3 sec2 x
−2
16. y = x2 tan(5x)
answer: 2x tan(5x) − 5x2 sec2 (5x)
8. f (x) = (cos x + 3 sin x)4 .
answer: 4(cos x + 3 sin x)3 (3 cos x − sin x) 1 + sin x2
17. y = .
√ 1 + cos (x2 )
9. y = sin x
1+cos x2
2 x 2 answer:
answer:
cos√ x 2x 1+cos x2 +sin x2
√
Solution. 44.2 q 1
0 2 2 0
0
− xq
q q
1+ 2
2 2
1+ −x 1+ x2 1+ x22 x2
q x =
x2 x2
=
2 2
1+ 2 1+ x2 1+ x2
x2 q
2 q2
1+ x2 + 2
x2 1+ x22 x2 1 + x2 +2 x2 + 4
= 2 = 32 = 32
1+ x2 x2 1 + 2
x2 1 + 2
x2 x2
Please note that this problem can be solved also by applying the transformation
x x x ±x2
q =q = 1
√ =√
1+ 2 x2 +2
±x x2 + 2 x2 + 2
x2 x2
√
before differentiating, however one must not forget the ± sign arising from x2 = ±x. Our original approach resulted in more algebra,
but did not have the disadvantage of dealing with the ± sign.
Solution. 44.3
Let u = cos x.
1
Then y = u2 .
dy dy du
Chain Rule: =
dx du dx
1 −1
= u 2 (− sin x)
2
1 1
= − sin x(cos x)− 2 .
2
28
Solution. 44.5
Let u = 1 + cos x.
Then y = u2 .
dy dy du
Chain Rule: =
dx du dx
= (2u)(− sin x)
= −2 sin x(1 + cos x)
= −2 sin x − 2 sin x cos x
= −2 sin x − sin(2x). (This last step is optional.)
Solution. 44.9
√
Let u= x.
Then y = sin u.
dy dy du
Chain Rule: =
dx du dx
1 −1
= (cos u) u 2
2
√
cos ( x)
= √ .
2 x
Solution. 44.15
dy 1 − 21 d
Chain Rule: = (sec(4x)) (sec(4x))
dx 2 dx
!
dy 1 d
Chain Rule: = p (sec(4x) tan(4x)) (4x)
dx 2 sec(4x) dx
!
1
= p (sec(4x) tan(4x))(4)
2 sec(4x)
2 sec(4x) tan(4x)
= p
sec(4x)
There are many ways to simplify this answer, including both of the following.
p
= 2 sec(4x) tan(4x).
3
= 2(sec(4x)) 2 sin(4x).
Solution. 44.16
dy d d
Product Rule: = (x2 ) (tan(5x)) + (tan(5x)) (x2 )
dx dx dx
dy
= (x2 )(−5 sec2 (5x) + (tan(5x))(2x)
dx
= 2x tan(5x) − 5x2 sec2 (5x).
29
Solution. 44.17
d
d
dy 1 + cos x2 dy (1 + sin x2 ) − (1 + sin x2 ) dx (1 + cos x2 )
Quotient Rule: =
dx (1 + cos (x2 ))2
d d
By the Chain Rule, dx (1 + cos x2 ) = −2x sin x2 and dx (1 + sin x2 ) = 2x cos x2 .
dy (1 + cos x2 )(2x cos x2 ) − (1 + sin x2 )(−2x sin x2 )
=
dx (1 + cos (x2 ))2
2x cos x2 + 2x cos2 x2 + 2x sin x2 + 2x sin2 x2
=
(1 + cos (x2 ))2
2
2x(cos x + sin x2 ) + 2x(cos x2 + sin x2 )
2 2
=
(1 + cos (x2 ))2
dy 2x + 2x(cos x2 + sin x2 )
=
dx (1 + cos (x2 ))2
2x(1 + cos x2 + sin x2 )
= .
(1 + cos (x2 ))2
Problem 45. Differentiate. The answer key has not been proofread, use with caution.
1
1. f (x) = sin(−5x). 4. f (x) = e x . 1+x
1 answer:
x2
7. f (x) = ln(1 + x3 )
answer: −5 cos(−5x) = −5 cos(5x) answer: − e x
1
√
answer: (−200x + 400) 4x − x2
99
9. f (x) = 1 − 2x.
3. f (x) = (2x − 3)4 (x2 + x + 1)5 .
answer: −(1 − 2x) 2
−1
−7 − 12x + 28x2 (−3 + 2x)3 1 + x + x2 answer:
4
r
2 3 2 6 x2 + 1
4. f (x) = (x + 1) (x + 2) . 10. f (x) = .
x2 + 4
24x + 18x3 1 + x2 2 + x2 answer: x2 +4 x2 +4
2 5 answer: 3x 2
2 x2 +1
− 1
1
6. f (x) = . 1
1 + x2 12. f (x) = .
(1+x2 )2
−2x answer: (1 + sec x)2
(1+cos (x))3 (1+cos (x))3
answer: =
−2 cos (x) sin (x) − sin (2x)
3
x2 + 1
7. f (x) = . √
3
x2 − 1 13. f (x) = 1 + tan x.
30
31
Problem 49. Find the derivative of the following functions.
Problem Collection All Techniques 6.6
1 1 1 3 1 answer: 9e−3x
answer: 1 − + = 2x 2 2 − 2
2 (x+1)2 (−x+1)2 answer: ex x 2 + 1 ex x−1
1−x2 4
1−x 2 3. f (x) = e−3x .
ln 7. f (x) = . 5. f (x) = e
8 cos (2x)
x
answer: 8 cot(2x) csc2 (2x) =
1+x 1
√
sin3 (2x)
−8 −e−x +ex 2. f (x) = cot(2x).
answer: 3 −1 −3 −1 −4
e−x +ex answer: 2ex x + ex x
answer: 25 sin (5x)
ex + e−x
6. f (x) = 4. f (x) = e x . 1. f (x) = sin(−5x).
1
ex − e−x
Problem 48. Compute the second derivative.
answer: 4((sin2 x + x)3 + x)3 (3(sin2 x + x)2 (2 sin x cos x + 1) + 1) − 1
−2x x2 +b2 2 −2a
answer: !3
1
16. f (x) = x + (x + sin2 x)3 . x2 +b2 2 +ax
4
p
− 1 π cos (tan (πx)) sin sin(tan (πx))
8. f (x) = ax + x2 + b2 .
answer: 2 p
p
sin (tan (πx)) cos2 (πx)
−2
−2 cos t cos (sin t)
answer:
sin(tan(πx)). 15. f (x) = cos
p sin3 (sin t)
answer: −12 cos x sin2 x sin (sin3 x) cos3 (sin3 x) 7. f (t) = cot2 (sin t).
14. f (x) = cos4 (sin3 x).
− 1
− 1 x2 +2 x 2
answer: 2
(x2 +4)2 x2 +4
answer: pr(2r sin (rx) + n)p−1 cos (rx)
x2 + 4
13. f (x) = (2r sin(rx) + n)p . . 6. f (x) =
x
r
1
−
1
!1 2 1
!− 1
1
! 3
2 1 2 16 sin (2x) − cos (2x)+1
answer: 1 x 2 + x + x x2 + x 1 x− 2 + 1 + 1 answer:
2 2 2 (cos (2x)+1)2 cos (2x)+1
1 + cos(2x)
x. x+ x+ 12. f (x) = . 5. f (x) =
q
1 − cos(2x)
√
r 4
! !− 1
1 1 2 answer: −x−1 cos (x−1 ) + sin (x−1 )
answer: 1 + 1 x− 2 x2 + x
2 4
x
x. x+ 11. f (x) = 4. f (x) = x sin .
1
√
q
answer: cos x cos (sin x) cos (sin (sin x)) x
answer: 4 sin
cos3 x
10. f (x) = sin(sin(sin x)). 3. f (x) = sec2 x + tan2 x.
2
−45(−3x + 1)4 + 6x (−3x + 1)5 + x2
Using computer algebra: 2m sin (mx)
answer:
2 (cos (mx))3
(−3645x4 + 4860x3 − 2430x2 + 546x − 45) (−3x + 1)5 + x2
Using computer algebra full expansion:
answer: −215233605x14 + 1004423490x13 − 2176250895x12
2. f (x) = sec2 (mx).
+2903793624x11 − 2666357595x10 + 1782098820x9
−893713176x8 + 341444160x7 − 99805041x6
+22199676x5 − 3697470x4 + 447132x3 answer: 2 sec2 (2x) cos(tan 2x)
−37125x2 + 1896x − 45
9. f (x) = x2 + (1 − 3x)5 . 1. f (x) = sin(tan(2x)).
3
Problem 47. Differentiate.
cos (kx)+kx sin (kx) answer: −3x2 sin 2 + x3
answer:
(cos (kx))2
16. f (x) = x sec(kx).
14. f (x) = cos(2 + x3 ).
answer: x−2 sin x−1 sin x2 + 2x cos x−1 cos x2
x
−2
answer: 1 (1 + tan x) 3 sec2 x
15. f (x) = cos
3
sin(x2 ).
1
32
answer: y = 7 x − 3 answer: y = −2
2 2
2. sin(x + y) = 2x − 2y, (π, π) .
3. x2 + xy + y 2 = 3, (1, −2) (ellipse).
answer: y = 1 x
2
perbola).
answer: 1 x + 2 π
3 3
4. x2 + 2xy − y 2 + x = 2, (1, 2) (hy- . 2, 4 1. y sin(2x) = x cos(2y),
π π
the tangent line to the curve at the given point. The answer key has not been proofread, use with caution.
Problem 51. Verify that the coordinates of the given point satisfy the given equation. Use implicit differentiation to find an equation of
dy y 3 −6x2 −2xy −2xy cos x2 +sin y 2
answer: = dy
dx −3xy 2 +x2 answer: =
dx −2xy cos y 2 +sin x2
18. 2x3 + x2 y − xy 3 = 2.
10. y sin x2 = x sin y 2 .
dy
dy −4x3 y+3y 2 −5x4 answer: = tan x tan y
answer: = dx
dx 3y 2 +x4 −6yx
9. 4 cos x sin y = 1.
17. x4 (x + y) = y 2 (3x − y).
dy y sin(xy)
answer: = −
dx cos y+x sin(xy)
dy −(sec (−y+x))2 x4 −2(sec (−y+x))2 x2 −(sec (−y+x))2 −2yx
answer: =
dx −(sec (−y+x))2 x4 −2(sec (−y+x))2 x2 −(sec (−y+x))2 −x2 −1 8. cos(xy) = 1 + sin y.
1 + x2
dy y sin x+2x
.
answer: =
16. tan(x − y) =
y dx cos x−2y
dy cos (xy)y+y sin x
answer: =
7. y cos x = x2 + y 2 .
dx − cos (xy)x+cos x
dy 4x3 y−2xy 3
answer: =
15. y cos x = 1 + sin(xy). dx 5y 4 +3x2 y 2 −x4
dy −y cos x−sin y
6. y 5 + x2 y 3 = 1 + x4 y.
answer: =
dx x cos y+sin x
dy −5x4 −4x3 y+3y 2
answer: =
14. x sin y + y sin x = 1. dx x4 −6xy−3y 2
y
5. x4 (x + y) = y 2 (3x − y).
2xy− √
dy 2 xy
= x
dx √ −x2
answer: 2 xy dy y 3 −6x2 −2xy
answer: =
4xyx2 y+4xy−y dx −3xy 2 +x2
=
−2x4 y−2x2 +x
4. 2x3 + x2 y − xy 3 = 2.
xy = 1 + x2 y. 13.
√
dy −2x−y
answer: =
dx x−2y
−1
dy − 1 (y+x) 2 +8xy 2
answer: = 2
dx 1
1 (y+x)− 2 −8x2 y
3. x2 + xy − y 2 = 4.
2
q
dy y
answer: = −2
12.
dx x
x + y = 1 + x2 y 2 .
√
y
−y 2 +y sec2
2. 2 x + y = 3.
dy
answer: = x
√
dx
√
y 2 +x sec2 x
y 2
dy
answer: = − x2
y dx y
11. tan = x + y. 1. x3 + y 3 = 1.
x
as a function of x and y by implicit differentiation. The answer key has not been proofread, use with caution. dx Problem 50. Express
dy
Implicit Differentiation 6.7
0 − sin y − y cos x
answer: y =
x ln x 3
4.
sin x + x cos y √
1 1
answer: · 1+
8. Find y 0 if x sin y + y sin x = 4.
x− 1 x2
x
0 −4x − 1 − y 2
answer: y = = −2x2 −1
x
x x 3. ln x −
1
7. Find y 0 if 2x2 + x + xy = 1.
q
− 1
x2 +1 1 2
2 2 e · x +1 2 · 2x
answer: 3 sin (2x) · cos(2x) · 2 = 6 sin (2x) cos(2x)
2
answer: q
x2 +1
= xe
q
6. sin3 (2x)
x2 +1
x x
2. e
answer: − sin e ·e x2 +1
√
x cos x − 2 sin x
answer:
5. cos(ex )
x3
1 1 1
x2
answer: p ln x + p
1.
3 3 x2 3 2
x
sin x
4 4 √
5. x2 + y 2 = (2x2 + 2y 2 − x)2 , (0, 12 ). 7. 2(x2 + y 2 )2 = 25(x2 − y 2 ), (3, 1). 9. x 3 + y 3 = 10 at (−3 3, 1).
2 13 13
answer: y = x + 1 answer: y = − 9 x + 40 3x + 10 answer: y =
√
2 2 √
6. x 3 + y 3 = 4, (−3 3, 1). 8. y 2 (y 2 − 4) = x2 (x2 − 5), (0, −2). 10. x2 y 3 + x3 − y 2 = 1 at (1, 1).
3
1 x+4
answer: y = √ answer: y = −2 answer: y = −5x + 6
Solution. 51.1
y
1 x
π π
First we verify that the point (x, y) = 2, 4 indeed satisfies the given equation:
π
y sin(2x)|x= π2 ,y= π4 = sin π = 0 left hand side
π 4 π
x cos(2y)|x= π2 ,y= π4 = cos = 0 right hand side
2 2
so the two sides of the equation are equal (both to 0) when x = π2 and y = π4 .
dy π
Since we are looking an equation of the tangent line, we need to find dx π - that is, the derivative of y at the point x = 2,
|x= π
2 ,y= 4
π
y = 4 . To do so we use implicit differentiation.
d
y sin(2x) = x cos(2y) dx
dy d d
sin(2x) + y (sin(2x)) = cos(2y) + x (cos(2y))
dx dx dx
dy dy
sin(2x) + 2y cos(2x) = cos(2y) − 2x sin(2y)
dx dx
dy π π
(sin(2x) + 2x sin(2y)) = cos(2y) − 2y cos(2x) Set x = 2,y = 4
dx π π π
dy
sin π + π sin = cos − cos π
dx |x= π2 ,y= π4 2 2 2
dy
π = − π2 cos π
dx |x= π2 ,y= π4
dy 1
= .
dx |x= π2 ,y= π4 2
π π
Therefore the equation of the line through x = 2,y = 4 is
π 1 π
y− = x−
4 2 2
1
y = x.
2
33
Solution.
π
LS = 1 · arctan 1 + 1 · arctan 1 RS = .
2
π π
=1· +1·
4 4
π
=
.
2
The fact that the left side equals the right side when we plug in x = 1 and y = 1 means that the point (1, 1) is on the graph of the
relation.
dy
2. Find dx in terms of x and y.
1
dy 2 + arctan 1
= − 1+1
1
dx 1+12 + arctan 1
= −1.
Now use the point (1, 1) to find an equation for the tangent line.
y − 1 = (−1)(x − 1)
y = −x + 2.
34
dy
2. Find dx in terms of x and y.
dy dy 1
x2 + 2xy = −y 2 − 2xy − √
dx dx 1 − x2
dy 1
(x2 + 2xy) = − y 2 + 2xy + √
dx 1 − x2
1
dy y 2 + 2xy + √1−x 2
=− 2
.
dx x + 2xy
3. Find the equation of the tangent to the graph at each of the points you found in the first part.
dy
Solution. To find the slope of the tangent at ( 21 , 0), plug in x = 21 , y = 0 to the formula for dx .
(0)2 + 2( 21 )(0) + √ 1
dy 1−( 12 )2
=−
dx ( 12 )2 + 2( 12 )(0)
0 + 0 + √1 3
4
=− 1
4 +0
√
1
3
=− 2
1
4
2 4
= −√ ·
3 1
8
= −√ .
3
Now use the point ( 12 , 0) to find an equation for the tangent line.
8 1
y − 0 = −√ x−
3 2
8 4
y = −√ x + √ .
3 3
35
(− 21 )2 + 2( 12 )(− 12 ) + √ 1
dy 1−( 12 )2
=−
dx ( 12 )2 + 2( 12 )(− 12 )
1 1
4 − 2 + √1 3
4
=− 1 1
4 − 2
− 41 + √2
3
=−
− 14
− 14 + √2
3
= 1
4
1 2
=4 − +√
4 3
8
= −1 + √ .
3
Now use the point ( 12 , − 12 ) to find an equation for the tangent line.
1 8 1
y− − = −1 + √ x−
2 3 2
8 1 4 1
y = −1 + √ x+ − √ −
3 2 3 2
8 4
y = −1 + √ x− √ ,
3 3
and this is the equation of the other tangent line.
2. xtan x . x
tan x + (ln x) sec2 x answer: xtan x
0
sin x 0
(ln x) sin x (ln x) sin x 0 sin x sin x
Solution. 54.1. x = e =e (ln x sin x) = x + ln x cos x .
x
Problem 55. Differentiate.
3 x
1. 10x . answer: 3(ln 10)x2 (10)x
3
4. xx . answer: (ln(x))2 xx +x + xx +x−1 + (ln x)xx +x
x x x
3. xx . answer: xx (log (x) + 1) 6. (ln x)ln x . answer: ln (ln (x))x−1 (ln (x))ln (x) + x−1 (ln (x))ln (x)
2. A car drives along an elliptical track. The track can be modeled by the equation x2 + 5y 2 = 14, where x and y are measured in
kilometres of distance from the center of the track. As the car passes the point (3, 1), the x-coordinate is increasing at a rate of
1.5 km/min. How quickly is the y-coordinate changing at that point?
dt 10
= − 9 km/min. answer:
dy
36
3. Gravel is being dumped from a conveyor belt at a constant rate of 500 litres per minute. The gravel pile forms a cone with circular
base, the diameter of which remains equal to the hight of the cone at any given moment. Use related rates to approximate how
fast is the height of the pile increasing when it is 2 meters tall? dt 2π
answer: dh = 1 m/min.
Solution. 56.1 Let t denote time. Let R be the radius of the sphere. Let S be the surface area of the sphere. We are given that
dS
= −50 ,
dt
where the sign is negative because the bubble is decreasing. We are asked to find
dR
=?
dt
R and S are related via
S = 4πR2 differentiate with respect to time
dS dR
= 4π ∗ 2R ∗
dt dt
dR 1 dS
= .
dt 8πR dt
When S = 1000, we can find the corresponding value of R:
1000 = 4πR2
250
R2 =
rπ
250
R = .
π
Finally, we can compute: √
dR 1 10π
= r ∗ (−50) = − .
dt |S=1000 250 |{z} 8π
8π = dS
dt
π
| {z }
=R when S=1000
dx
Solution. 56.2. Let t denote time, and let t0 be the point in time in which the measurements take place. We are given that dt |t=t0 =
dy
1.5km/min, x|t=t0 = 3, y|t=t0 = 1. The problem asks us to find dt |t=t0 .
Compute:
d
x2 + 5y 2 = 14 apply
dt
dx dy
2x + 10y = 0 fix time t = t0
dt dt
dy
2 · 3 · 1.5 + 10 · 1 · = 0
dt |t=t0
dy
10 = −9
dt |t=t0
dy 9
= − .
dt |t=t0 10
9
The measurement unit of dy is km, the measurement unit of t is minutes, therefore the answer is − 10 km/min.
37
Solution. 56.3.
At time t (minutes) let h be the height of the pile (m), let r be the radius of the
base (m), and let V be the volume of the pile (m3 ).
We are given that dV 1 3
dt = 500L/min = 2 m /min. We are also given the diameter
is equal to the height, so 2r = h.
We are asked to find dhdt when h = 2. h
The formula for the volume of a cone is
1 2
V = πr h.
3
h
r
Since r = 2 we obtain
1 h2 1
V = π h= πh3
3 4 12
Differentiating with respect to t then gives
dV 1 dh dh 4 dV
= πh2 ⇒ = 2
dt 4 dt dt h π dt
dV
Substituting dt = 12 m3 /min, and h = 2 we obtain
dh 4 1 1
= · = m/min
dt 4π 2 2π
dV dx
Problem 57. 1. If V is the volume of a cube with edge length x and the cube expands as time passes, find dt in terms of dt .
dt
answer: 3x2 dx
2. Each side of a square is increasing at a rate of 1cm/s. At what rate is the area of the square increasing when the area of the
square is 9 cm2 ? answer: 6cm2 /s
3. The radius of a ball is increasing at a constant rate of 5mm/s. At a point in time we measure the radius of the ball to be 5cm.
• How fast is the volume increasing at the time of our observation?
• How fast is the surface area increasing at the time of our observation?
4. At a point in time, we measure the surface area of a ball to be increasing at a rate of 5cm2 /s and the radius of the ball to be 5cm.
• How fast is the volume increasing at the time of our observation?
• How fast is the radius increasing at the time of our observation?
5. At a point in time, we measure the volume of a ball to be increasing at a rate of 5cm3 /s and the radius of the ball to be 5cm.
• At the time of observation, how fast is the bottom end of the ladder sliding away from the wall
• At the time of observation, how fast is the midpoint of the ladder moving away from the wall?
• At the time of observation, how fast is the midpoint of the ladder moving downwards towards the ground?
7. A street light is mounted at the top of a 4m tall pole. A woman 160 cm tall walks away from the pole at a speed of 5km/h along a
straight path. How fast is the tip of her shadow moving when she is 10m from the pole?
8. A ship is pulled into a dock. On the dock, the rope is pulled by a pulley that is 1m lower than the mooring point on the ship. If the
rope is pulled in at a constant rate of 10cm/s, how fast is the mooring point approaching the dock when it is 10m from the dock?
9. A Ferris wheel with a radius of 10m is rotating at a rate of one revolution every 2 minutes. How fast is a riding rising when his
seat is 16 m above ground level?
10. The minute hand on a watch is 10mm long and the hour hand is 5mm long. How fast is the distance between the tips of the hands
changing at two o’clock? √
21π mm/h
42
answer: (not proofread) − 55
38
Solution. 57.1. The volume V is given by V = x3 , therefore
dV d 3 dx
= x = 3x2 .
dt dt dt
Solution. 57.4 Let S denote the surface area and r the radius of the ball. Then S = 4πr2 . Let the point of time be t0 . We are given
that dS 2
dt |t=t0 = 5cm /s. On the other hand,
dS d
4πr2 = 8πr dr
dt = dt dt
1 dS
8π dt
Solution. 57.10 Let the angle between the two arrows be θ. The cosine law states that for a triangle with angle θ and sides a, b, c we
have that c2 = a2 + b2 − 2ab cos θ (where c is the length of the side opposite to the angle θ).
Then by the cosine law, the distance between the tips of the two hands is 1
p √
y = 52 + 102 + 2 · 5 · 10 cos θ = 125 + 100 cos θ.
The short hand makes 1 full revolution every 12 hours, and the long hand makes 1 full revolution every 1 hour. Therefore the angle
θ measured from the small hand to the long hand changes at the (constant) rate of 11 12 revolutions per hour, or what is the same, at the
dθ 11 11
rate dt = 12 (2π) = 6 π.
The problem asks us to compute dydt at two o’clock, i.e., at t = 2. This is a straightforward computation using the chain rule:
dy 1 −100 sin θ dθ π dθ 11
= √ at 2 o’clock θ = 3 and dt = 6 π
dt 2 125 + 100 cos θ dt
55
√
The measurement unit of speed is mm/hour, so the distance is changing at the rate of − 42 21 mm/hour.
39
Problem 59. 1. We have that f is continuous in [0, 10] and differentiable in (0, 10). If f (0) = 0, f (10) = 10 and |f 0 (x)| ≤ 1, how
large can f (x) be for x in the interval [0, 10]?
answer: 10
2. We have that f is continuous in [0, 10] and differentiable in (0, 10). If f (0) = −3, f (2) = −13 and |f 0 (x)| ≤ 5, what is the
smallest possible value of f (1)?
answer: −8
Problem 60. Use the Intermediate Value theorem and the Mean Value Theorem/Rolle’s Theorem to prove that the function has exactly
one real root.
1. f (x) = x3 + 4x + 7.
2. f (x) = x3 + x2 + x + 1.
Solution. 60.1. f (−2) = −9 and f (1) = 12. Since f (x) is continuous and has both negative and positive outputs, it must have a zero.
In other words, for some c between −2 and 1, f (c) = 0. If there were solutions x = a and x = b, then we would have f (a) = f (b), and
Rolle’s Theorem would guarantee that for some x-value, f 0 (x) = 0. However, f 0 (x) = 3x2 + 4, which always positive and therefore
is never 0. Therefore there cannot be 2 or more solutions.
The above can be stated informally as follows. Note that f 0 (x) = 3x2 + 4, which is always positive. Therefore, the graph of f is
increasing from left to right. So once the graph crosses the x-axis, it can never turn around and cross again, so there can only be a single
zero (that is, a single solution to f (x) = 0).
= cos3 35 + sin 5 − 15 ≤ 2 − 15 = −13 < 0 (because cos a, sin b ∈ [−1, 1] for arbitrary a, b). Similarly
Solution. 60.3. f (5)
f (−5) = cos3 − 35 + sin(−5) + 15 ≥ 15 − 2 > 0. Therefore by the Intermediate Value Theorem f (x) = 0 has at least one solution
x
2. f (x) = 5 + 4x − 2x3 , x ∈ [−1, 1]. 7. f (x) = , x ∈ [0, 3].
x2 − x + 1
3 9 min = f − 3 9 answer: fmax = f () =, fmin = f () =
6 = 8 √6 + 5, f 6 = 5 − 8 √6 answer: fmax = f
√ √
p
3 2 8. f (t) = t 4 − t2 , x ∈ [−1, 2].
3. f (x) = 2x − x − 20x + 1, x ∈ [−4, 3].
answer: fmax = f () =, fmin = f () =
3 27
− 5 = 602 , fmin = f (−4) = −63
answer: fmax = f
√
3
9. f (t) = t(8 − t), x ∈ [0, 8].
4. f (x) = 3x4 − 4x3 − 12x2 + 1, x ∈ [−2, 3]. answer: fmax = f () =, fmin = f () =
40
Problem 62. A particle moves in such a way that, after t seconds, it is s(t) = ln 2 − t + t2 m to the right of the origin.
Solution.
−1 + 2t
s0 (t) = .
2 − t + t2
Set s0 (t) = 0.
−1 + 2t
=0
2 − t + t2
−1 + 2t = 0
1
t= .
2
Therefore the position function has a critical number at t = 21 . The parabola 2 − t + t2 has a global minimum at t = 12 , and the
function, so ln 2 − t + t also has a global minimum at t = 12 . The minimum value
2
natural logarithm
functionisan increasing
2
is s( 12 ) = ln 2 − 12 + 21 = ln 74 ≈ 0.5596 m.
Solution.
−1 + 2t
v(t) = s0 (t) = .
2 − t + t2
(2 − t + t2 )(2) − (−1 + 2t)(−1 + 2t)
a(t) = s00 (t) =
(2 − t + t2 )2
(4 − 2t + 2t ) − (1 − 4t + 4t2 )
2
=
(2 − t + t2 )2
3 + 2t − 2t2
= .
(2 − t + t2 )2
3 + 2( 12 ) − 2( 12 )2
1 1
Plug in t = : a =
2 2 7 2
4
1
3+1− 2
= 49
16
7
2
= 49
16
8
= .
7
8
Therefore the particle is accelerating at a rate of 7 ≈ 1.1429 m/s2 when it is closest to the origin.
Solution. The natural logarithm function is defined for all positive input values. The formula y = 2 − t + t2 is always positive.
To see this, note that its graph is a parabola with discriminant b2 − 4ac = (−1)2 − 4(1)(2) = −7, which is negative. This means
that the graph of y = 2 − t + t2 never touches the x-axis, and hence it is either always positive or always negative. Since y(0) = 2
is positive, this means all values are positive. Therefore ln(2 − t + t2 ) is defined for all input values t.
41
x 1
1. f (x) = . 4. f (x) = x + .
1 + x2 x
answer: x = 1, local & global max, x = −1, local and global min
answer: x = −1, local max, x = 1, local min
2. f (x) = x3 − x2 − x − 1.
x − 12
3
answer: x = − 1 , local max, x = 1, local min
5. f (x) = .
3. f (x) = 2x3 − x2 − 20x + 1. x2 − 2x + 47
3 2 2
answer: x = − 5 , local max x = 2, local min answer: x = − 1 , local and global min, x = 3 , local and global max
7.2.3 Optimization
Problem 64. 1. Find the dimensions of a rectangle with area 1000 m2 whose perimeter is as small as possible.
2. A box with an open top is to be constructed from a square piece of cardboard, 1m wide, by cutting out a square from each of the
four corners and bending up the sides. Find the largest volume that such a box can have.
3. A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible volume of such a cylinder.
4. A wedge of radius 2 (depicted below) is folded into a cone cup. The volume varies depending on the angle of the wedge. Find the
A B
O
r
maximal possible volume of the cone cup and the angle of the wedge for which this maximal volume is achieved.
Problem 65. 1. What is the x-coordinate of the point on the hyperbola x2 − 4y 2 = 16 that is closest to the point (1, 0)?
answer: x = 4
2. What is the x-coordinate of the point on the ellipse x2 + 4y 2 = 16 closest to the point (1, 0)?
3
answer: x = 4
3. A rectangular box with a square base is being built out of sheet metal. 2 pieces of sheet will be used for the bottom of the box,
and a single piece of sheet metal for the 4 sides and the top of the box. What is the largest possible volume of the resulting box
that can be obtained with 36m2 of metal sheet? answer: 12 cubic meters.
4. Recall that the volume of a cylinder is computed as the product of the area of its base by its height. Recall also that the surface
area of the wall of a cylinder is given by multiplying the perimeter of the base by the height of the cylinder.
A cylindrical container with an open top is being built from metal sheet. The total surface area of metal used must equal 10m2 .
Let r denote the radius of the base of the cylinder, and h its height. How should one choose h and r so as to get the maximal
possible container volume? What will the resulting container volume be?
Solution. 65.1 p
The distance function between an arbitrary point (x, y) and the point (1, 0) is d = (x − 1)2 + (y − 0)2 . On the other hand, when
2
the point (x, y) lies on the hyperbola we have y 2 = x −16
4 . In this way, the problem becomes that of minimizing the distance function
r
p x2 − 16
dist(x) = (x − 1)2 + y 2 = (x − 1)2 + .
4
0
This is a standard optimization problem: we needr to find the critical endpoints, i.e., the points where dist = 0. As the square root
x2 − 16
function is an increasing function, the function (x − 1)2 + achieves its minimum when the function
4
x2 − 16
l = dist2 = (x − 1)2 +
4
42
does. l is a quadratic function of x and we can directly determine its minimimum via elementary methods. Alternatively, we find the
critical points of l:
l0 = 0
x
2(x − 1) + = 0
2
5
x−2 = 0
2
4
x = .
5
√
On the other hand, x2 = 16 + 4y 2 and therefore |x| ≥ 16 = 4. Therefore x ∈ (−∞, −4] ∪ [4, ∞). As x = 54 is outside of the allowed
range, it follows that our function either attains its minimum at one of the endpoints ±4 or the function has nop
minimum at all. It is clear
however that as x tends to ∞,q so does dist. Therefore dist attains its minimum
q for x = 4 or −4 and y = ± (±4)2 − 16 = 0. Direct
2 2
check shows that dist|x=4 = (4 − 1)2 + 4 −16 4 = 3 and dist|x=−4 = (−4 − 1)2 + 4 −16 4 = 5 so our function dist has a minimal
value of 3 achieved when x = 4, which is our final answer. Notice that this answer can be immediately given without computation by
looking at the figure drawn for 65.1. Indeed, it is clear that there are no points from the hyperbola lying inside the dotted circle centered
at (1, 0). Therefore the point where this circle touches the hyperbola must have the shortest distance to the center of the circle.
Solution. 65.3 Let B denote the area of the base of the box, equal to the area of the top. Let W denote the area of the four walls of the
box (the four walls are all equal because the base of the box is a square). Then the surface area S of material used will be
S= 2B
|{z} + |{z} B = 3B + 4W
4W + |{z} .
two pieces for the bottom 4 walls the box lid
Let x denote the length of the side of the square base and let y denote the height of the box. Then
B = x2
and
W = xy .
As the surface area S is fixed to be 36 square meters, we have that
S = 3B + 4W = 36 = 3x2 + 4xy .
√
As y is positive, the above formula shows that 3x2 ≤ 36 and so x ≤ 12. Let us now express y in terms of x:
3x2 + 4xy 36 =
4xy 36 − 3x2
=
36 − 3x2
y = .
4x
The problem asks us to maximize the volume V of the box. The volume of the box equals the area of the base times the height of the
box:
(36 − 3x2 ) 2 36x − 3x3
V = B · y = yx2 = x = .
4x 4
As x is non-negative, it follows that the domain for x is:
√
x ∈ [0, 12] .
To maximize the volume we find the critical points, i.e., the values of x for which V 0 vanishes:
0
36x − 3x3
0 = V0 =
4
36 − 9x2
0 =
4
9x2 = 36
x2 = 4
x = ±2
As x measures length, x = −2 is not possible
√ (outside of the domain for x). Therefore the only critical point is x = 2. Direct check
shows that at the endpoints x = 0 and x = 12, we have that V = 0. Therefore the maximal volume is achieved when x = 2:
36(2) − 3(2)3
Vmax = V|x=2 = = 12 .
4
43
44
ex − e−x
4. f (x) =
ex + e−x
x
y
q
For f (x) = q 1 : 2 −x2 +1+1
For f (x) = q :
−x2 +1+1
−x2 +1+1
y-intercept: x = 1 , no x intercept
2 y-intercept: x = 3 , no x intercept
no asymptotes 2
answer: answer: no asymptotes
decreasing on [−1, 0], increasing on [0, 1]
increasing on [−1, 0], decreasing on [0, 1]
global and local min at x = 0, global and local max at x = ±1.
global and local max at x = 0, global and local min at x = ±1.
concave up on [−1, 1]
concave down on [−1, 1]
no inflection points
no inflection points
The two functions are plotted simultaneously in the x, y-plane. Indicate which part of the graph is the graph of which function.
−x2 + 1 + 1 −x2 + 1 + 1
3. f (x) = √ , f (x) = √
x
2 −x2 + 1 + 1 1
√
y
y-intercept: 3
2
horizontal asymptote: y = 2, vertical: none
√ √ √ √
increasing on 3− 3 , 3+ 3 , decreasing on −∞, 3− 3 ∪ 3+ 3 , ∞
2 2 2 2
answer: √ √
local and global min at x = 3− 3 , local and global max at x = 3+ 3
2 2
concave up on 0, 3 ∪ (3, ∞), concave down (−∞, 0) ∪ 3 , 3
2 2
inflection points at x = 0, x = 3 , x = 3
2
x2 − 3x + 3
x
2. f (x) =
2x2 − 5x + 92
y
y-intercept: 1 . x-intercept: − 1
2 2
Horizontal asymptote: y = 0, vertical:
√ none √
local and global min at x = −1− 3 , local and global max at x = −1+ 3
2 √ √ 2 √ √
answer:
Intervals of decrease: −∞, −1− 3 ∪ −1+ 3 , ∞ , intervals of decrease −1− 3 , −1+ 3
2
2 2 2
Concave down on (−∞, −2) ∪ − 1 , 1 , concave up on −2, − 1 ∪ (1, ∞)
2 2
Inflection points at: x = −2, x = − 1 , x = 1
2
x +x+1
1. f (x) = 2
x
x + 12
y
Label all relevant points on the graph. Show all of your computations.
• points of inflection. • intervals of increase and decrease,
• horizontal and vertical asymptotes,
• intervals of concavity,
• x and y intercepts of f ,
• local and global minima, maxima, • the implied domain of f ,
45
x2 + 2x + 4
8. f (x) =
x+1
√
y-intercept: none, x-intercepts: −3∓ 5
2
horizontal asymptote: y = 1, vertical: x = −2 and x = 0
answer: always decreasing
no local/global minima/maxima
concave down on (−∞, −2) cup (−1, 0), concave up on (−2, −1) ∪ (0, ∞)
inflection point at x = −1
x2 + 2x
7. f (x) =
x2 + 3x + 1
x
y
−x + 1
6. f (x) = ln
x+1
x
y
e−x + ex
5. f (x) =
−e−x + ex
x
y
inflection points at x = −4, x = −1, x = 2
concave up on (−4, −1) cup (2, ∞), concave down (−∞, −4) ∪ (−1, 2)
local and global min at x = −1 − 3, local and global max at x = −1 + 3
√ √ answer:
increasing on −1 − 3, −1 + 3 , decreasing on −∞, −1 − 3 ∪ −1 + 3, ∞
horizontal asymptote:
√ √
y = 0, vertical: none √ √
4
y-intercept: 1 , x-intercept: −1
Solution. 66.2
Domain. We have that f is not defined only when
√
we have division
√
by zero, i.e., if x2 − 3x + 3 equals zero. However, the roots
2 3± 32 −4·3 3± −3
of x − 3x + 3 are not real numbers: they are 2 = 2 , and therefore x2 − 3x + 3 can never equal zero. Alternatively,
completing the square shows that the denominator is always positive:
2
3 9 9 3 3
x2 − 3x + 3 = x2 − 2 · x + − + 3 = x − + >0
2 4 4 2 4
Therefore the domain of f is all real numbers.
2 · 02 − 5 · 0 + 92 9
3
x, y-intercepts. The y-intercept of f equals by definition f (0) = 2
= 2 = . The x intercept of f is those values
0 −3·0+3 3 2
of x for which f (x) = 0. The graph of f shows no such x, and that is confirmed by solving the equation f (x) = 0:
f (x) = 0
2x2 − 5x + 92
= 0 Mult. by x2 − 3x + 3
x2 − 3x + 3
9
2x2 − 5x + = 0
2
√
q
5 ± 25 − 4 · 2 · 92 5 ± −9
x1 , x2 = = ,
4 4
√
so there are no real solutions (the number −9 is not real).
Asymptotes. Since f is defined for all real numbers, its graph has no vertical asymptotes. To find the horizontal asymptote(s), we
need to compute the limits lim f (x) and lim f (x). The two limits are equal, as the direct computation below shows:
x→∞ x→−∞
46
√ √ √ √
The points x1 , x2 split the real line into three intervals: −∞, 3−2 3
, 3−2 3 , 3+2 3 and 3+2 3 , ∞ , and each of the factors
of (1) has constant sign inside each of the intervals. If we choose√ x to be a very negative√
number,
√
it follows that −(x − x1 )(x − x2 )
is a negative, and therefore f (x) is negative for x ∈ (−∞, 2 ). For x ∈ ( 2 , 2 ), exactly one factor of f 0 changes sign
0 3− 3 3− 3 3+ 3
and √therefore f 0 (x) is positive in that interval; finally only one factor of f 0 (x) changes sign in the last interval so f 0 (x) is negative on
( 3+2 3 , ∞).
Our computations can be summarized in the following table.
Interval f 0 (x) f (x)
√
−∞, 3−2 3
− &
√ √
3− 3 3+ 3
2 , 2 + %
√
3+ 3
2 , ∞ − &
√
Local and global minima and maxima. The table above shows that f (x) changes from decreasing to increasing at x = x1 =
3− 3
2 and therefore f has a local minimum at that point. The table also shows that f (x) changes from increasing to decreasing at
√
x = x2 = 3+2 3 and therefore f has a local maximum at that point. The so found local maximum and local minimum turn out to be
global: there are two things to consider here. First, no other finite point is critical and thus cannot be maximum or minimum - however
this leaves out the possibility of a maximum/minimum “at infinity”. This possibility can be quickly ruled out by looking at the graph of
f . To do so via algebra, compute first f (x1 ) and f (x2 ):
√ 2 √
√ ! 2 3−2 3 − 5 3−2 3 + 92 √
3− 3 3
f (x1 ) = f = √ 2 √ =2−
2 3− 3 3− 3 3
2 − 3 2 + 3
√ 2 √
√ ! 2 3+ 3
−5 3+ 3
+ 92 √
3+ 3 2 2 3
f (x2 ) = f = √ 2 √ =2+ .
2 3+ 3 3+ 3 3
2 − 3 2 + 3
On the other hand, while computing the horizontal asymptotes, we established that lim f (x) = 2. This implies that all x sufficiently
x→±∞
far away from x = 0, we have that f (x) is close to 2. Therefore f (x) is larger than f (x1 ) and smaller than f (x2 ) for all sufficiently far
away from x = 0. This rules out the possibility for a maximum or a minimum “at infinity”, as claimed above.
Intervals of concavity. The intervals of concavity of f are governed by the sign of f 00 . The second derivative of f is:
!0
00 0 0 −x2 + 3x − 32
f (x) = (f (x)) = 2
(x2 − 3x + 3)
0 ! !0
2 3 1 2 3 1 second differentiation:
= −x + 3x − 2 + −x + 3x −
2 (x2 − 3x + 3) 2 (x 2 − 3x + 3)2 chain rule
0
!
x2 − 3x + 3
1 3
= (−2x + 3) 2 + −x2 + 3x − (−2) 3
(x2 − 3x + 3) ! 2 (x2 − 3x + 3)
1 (2x − 3)
+ 2x2 − 6x + 3 factor out (x2(2x−3)
= (−2x + 3) 2 3 −3x+3)2
2
(x − 3x + 3) (x 2 − 3x + 3)
(2x2 − 6x + 3)
(2x − 3)
= 2 −1 +
(x2 − 3x + 3) (x2 − 3x + 3) !
(2x − 3) − x − 3x + 3 + (2x2 − 6x + 3)
2
= 2
(x2 − 3x + 3) (x2 − 3x + 3)
2
(2x − 3)(x − 3x)
= 3
(x2 − 3x + 3)
(2x − 3)x(x − 3)
= 3
(x2 − 3x + 3)
When computing the domain of f , we established that the denominator of the above expression is always positive. Therefore f 00 (x)
changes sign when the terms in the numerator change sign, namely, at x = 0, x = 32 and x = 3.
Our computations can be summarized in the following table. In the table, we use the ∪ symbol to denote that the function is concave
up in the indicated interval, and ∩ to denote that the function is concave down.
47
Interval f 00 (x) f (x)
(−∞, 0) − ∩
(0, 32 ) + ∪
( 32 , 3) − ∩
(3, ∞) + ∪
Points of inflection. The preceding table shows that f 00 (x) changes sign at 0, 32 , 3 and
therefore
the points of inflection are located
at x = 0, x = 32 and x = 3, i.e., the points of inflection are (0, f (0)) = 0, 32 , 32 , f 32 = 32 , 2 , (3, f (3)) = 3, 52 .
We can ask our graphing device to use the so computed information to label the graph of the function. Finally, we can confirm
visually that our function does indeed behave in accordance with our computations.
y
√ √ √ √
3− 3 , 2 − 3 3+ 3 , 2 + 3
2 3 2 3
infl.: 0, 3 infl.: 3,2 infl.: 3, 5
2 2 2
x
Solution. 66.8
This problem is very similar to Problem 66.2. We recommend to the student to solve the problem first “with closed textbook”
and only then to compare with the present solution.
Domain. As f is a quotient of two polynomials (rational function), its implied domain is all x except those for which we get division
2 2
by zero for f . Consequently
√ the domain of f is all x for which x + 2x + 4 = 0. However, the polynomial x + 2x + 4 has no real
−2 ± 4 − 16 √
roots - its roots are = −1 ± −3, and therefore the domain of f is all real numbers. Alternatively, we can complete
2
the square: x + 2x + 4 = (x + 1) + 3 and so x2 + 2x + 4 is positive for all values of x.
2 2
0+1 1
x, y-intercepts. The y-intercept of f equals by definition f (0) = 2 = . The x intercept of f is those values of x for
0 +2·0+4 4
which f (x) = 0. We compute
f (x) = 0
x+1
= 0
x2 + 2x + 4
x+1 = 0
x = −1 ,
and the x-intercept of f is x = −1.
Asymptotes. The line x = a is a vertical asymptote when lim± f (x) = ±∞; as f is defined for all real numbers, this implies that
x→a
there are no vertical asymptotes.
The line y = L is a horizontal asymptote if lim f (x) exists and equals L. We compute:
x→±∞
(x + 1) x12 1 1
x + x2 0+0
lim f (x) = lim 2 1 = lim 2 4 = =0
x→∞ x→∞ (x + 2x + 4) 2
x
x→∞ 1 +
x + x2
1+0+0
Therefore y = 0 is a horizontal asymptote for f . An analogous computation shows that lim f (x) = 0 and so y = 0 is the only
x→±∞
horizontal asymptote of f .
Intervals of increase and decrease. The intervals of increase and decrease of f are governed by the sign of f 0 . We compute:
0
x+1
f 0 (x) = qutotient rule
x2 + 2x + 4
0
(x + 1)0 x2 + 2x + 4 − (x + 1) x2 + 2x + 4
= 2
(x2 + 2x + 4)
2
x + 2x + 4 − (x + 1)(2x + 2)
= 2
(x2 + 2x + 4)
x2 + 2x + 4 − 2x2 + 4x + 2
= 2
(x2 + 2x + 4)
2
−x − 2x + 2
= 2
(x2 + 2x + 4)
48
As x2 + 2x + 4 is positive, the sign of f 0 is governed by the sign of −x2 + 2x + 2. To find out where −x2 + 2x + 2 changes sign, we
compute the zeroes of this expression:
−x2 − 2x + 2 = 0
x2 + 2x − 2 = 0 √ use the quadratic formula
x1 , x2 = −1 ± 3 .
Problem 67. 1. Sketch the graph of y = x4 − 8x2 + 8 by determining the intervals of increase and decrease, finding the local mins
and maxes, determining where the graph is concave up and concave down, and plotting a few key points.
49
Local max at 0, local mins at 2 and -2. Concave down between − 4/3 and 4/3, and concave up otherwise.
Check your graph with a calculator or online graphing program. p p answer:
2. Sketch the graph of y = xx−1 2 −9 by graphing any vertical and horizontal asymptotes, finding the x- and y-intercepts, and then
√ √
2. Find the linearization of f (x) = 8 + x at a = 1 and use it to approximate 9.02.
6 6 6 300
answer: f (x) ≈ 3 + 1 (x − 1) = 1 x + 17 . Therefore 9.02 ≈ 901 ≈ 3.003333
√
√
3
√
3
3. Find the linearization of f (x) = 8 + x at a = 0 and use it to approximate 7.97.
12 400
√ 3 7.97 ' 799 = 1.9975
3 8 + x ≈ 1 x + 2. Therefore √ answer:
Solution. 68.6 Let f (x) = x2014 . We are looking to approximate (0.9999)2014 = f (0.9999). As f (1) = 12014 = 1 is easy to compute,
is makes sense to use linear approximation at a = 1 to approximate (0.9999)2014 . We have that
f 0 (x) = 2014x2013 .
Therefore
f (0.9999) ≈ 2014 · 0.9999 − 2013 = 1 · 0.9999 + 2013(0.9999 − 1) = 0.9999 − 2013 · 0.0001 = 0.9999 − 0.2013 = 0.7986
A computation with computer shows that 0.9992014 = 0.817577 . . . . While our approximation of 0.7986 is less than perfect, it is
within the same order of magnitude. We study techniques for estimating errors in linear approximations later.
9 Integration Basics
9.1 Riemann Sums
Problem 69. Estimate the integral using a Riemann sum using the indicated sample points and interval length.
Z 4
√
1. 8x + 1 dx. Use four intervals of equal width, choose the sample point to be the left endpoint of each interval.
0
0
17. f (x)dx ≈ 9 + 8x + 1. Thus answer: ∆x = 1 and f (x) =
√
√ Z 4
Z 6
1
2. dx. Use three intervals of equal width, choose the sample point to be the left endpoint.
0 x2 + 1
x2 +1 0 85
1 . Thus f (x)dx ≈ . answer: ∆x = 2 and f (x) =
Z 6 214
50
−0.5
Z
dx
3. . Use three intervals of equal width, choose the sample point to be the midpoint of each interval.
x2 + 1
−3.5
−3.5
5 x2 +1
f (x)dx ≈ ∆x (f (−3) + f (−2) + f (−1)) = = 0.8. 1 . Thus answer: ∆x = 1 and f (x) =
Z 4
−0.5
Z 2
dx 1
4. . Use ∆x = 2 and right endpoint sampling points.
0 1 + x + x3
20163 13 3 47 11 2
= 12197 ≈ 0.604920 8 + 1 + 8 + 1 answer: 1
Z 0
dx 2
5. . Use ∆x = 3 and left endpoint sampling points.
−2 1 + x + x2
819 3 13 7 3
= 1262 ≈ 1.540904 1 + 9 + 9 answer: 2
Z2
dx
6. . Use four intervals of equal width, choose the sample point to be the left endpoint of each interval.
1 + x3
0
0
2 2 1260 1+x3
f (0) + f (1) + f +f = ≈ 1.30873. f (x)dx ≈ ∆x 1 . Thus answer: ∆x = 0.5 and f (x) =
1 3 1649 Z2
Z0
dx
7. . Use four intervals of equal width, choose the sample point to be the right endpoint.
x4 +1
−2
0
2 2 6596 1+x3
f − + f (−1) + f − + f (0) = ≈ 1.303062. f (x)dx ≈ ∆x 1 . Thus answer: ∆x = 0.5 and f (x) =
3 1 8595 Z2
Solution. 69.1. The interval [0, 4] is subdivided into n = 4 intervals, therefore the length of each is ∆x = 1. The intervals are therefore
The problem asks us to use the left endpoints of each interval as sampling points. Therefore our sampling points are 0, 1, 2, 3. Therefore
the Riemann sum we are looking for is
√ √ √ √ √
∆x (f (0) + f (1) + f (2) + f (3)) = 1 · 8 · 0 + 1 + 8 · 1 + 1 + 8 · 2 + 1 + 8 · 3 + 1 = 9 + 17 ≈ 13.1231
1 2 3 4
Solution. 69.3. The interval [−3.5, −0.5] is subdivided into n = 3 intervals, therefore the length of each is ∆x = 1. The intervals are
therefore
[−3.5, −2.5], [−2.5, −1.5], [−1.5, −0.5] .
The problem asks us to use the midpoint of each interval as a sampling point. Therefore our sampling points are −3, −2, −1. Therefore
the Riemann sum we are looking for is
1 1 1
∆x (f (−3) + f (−2) + f (−1)) = 1 · + + = 0.8 .
10 5 2
51
y
−7 −5 −3 −1 x
2 2 2 2
9.2 Antiderivatives
Problem 70. Find all antiderivatives of the functions.
√ √ √
1. f (x) = 3 + π2 . 7. f (x) = x2.4 − 2x 3−1
. 13. g(x) =
1+
√
x+x
.
answer: x π + 3 + C
17 3
answer: 5 x 5 − 2 3x +C
x3
2 √
17 √ √3 answer: 2x 2 − 2x 2 + ln |x| + C
1 −1
2. f (x) = x − 5. 8
8. f (x) = . 14. f (t) = 3 sin t − 4 cos t.
− 5x + C
2
answer: x7
x2 3
answer: − 4 x−6 + C
answer: −3 cos t − 4 sin t + C
3. f (x) = x2 − 2x + 6. x+1
9. f (x) = . 15. f (θ) = sec2 θ.
3 x3
answer: x − x2 + 6x + C 2
3 answer: −x−1 − 1 x−2 + C
answer: tan θ + C
x(x + 1) 1
4. f (x) = . 10. f (x) = . 16. f (θ) = csc2 θ.
2 x
6 4 answer: ln |x| + C answer: − cot θ + C
answer: 1 x3 + 1 x2 + C
2
6. f (x) = 7x 7 + x− 7 .
4
5 − 4x3 + 2x6 2 + x cos x
12. f (x) = . 18. f (x) = .
9 3
x4 x
answer: 49 x 7 + 7 x 7 + C 3 3
9 3 answer: 2 x3 − 5 x−3 − 4 ln |x| + C answer: 2 ln |x| + sin x
Z3 Z2 Z1
x2 − 1 dx. (x − 1)(x2 + 1)dx. (1 + x2 )3 dx.
1. 3. 5.
−2 0 0
−2 4 3 2 0 3 7 5 0 35
3 3
1 x3 − x 3 = 20 answer: 1 x4 − 1 x3 + 1 x2 − x 2 = 4 answer: 1 x7 + 3 x5 + x3 + x 1 = 96 answer:
i h i h i h
Z2 Z1 2 Z2
x(x + 1)
3 2
1 4
2. 4x + 3x + 2x + 1 dx. 4. dx. 6. − dx.
2 x x2
1 −1 1
1 20 8 12 −1 15 1
answer: x4 + x3 + x2 + x = 26 1 x5 + 1 x4 + 1 x3 1 = 4 answer: answer: 4x−1 + ln x = ln 2 − 2
h i2 i h h i2
52
π
Z4 Z2 2 Z2
√ 1
7. x(1 + x)dx. 13. x+ dx. 20. csc2 θdθ.
x
1 1 π
1 4
15 5 3 answer: [] =
= 256 2x2 + 2x2 answer: answer: [] =
5 3
#4 "
Z2 3 π
Z4 r 1 Z4
6 14. x+ dx. 1 − cos2 θ
8. dx. x 21. dθ.
x 1 cos2 θ
0
1 answer: [] =
1 answer: [] =
answer: 2 6 x = 2 6
h √ √ i4 √
Z2 2
√ 1 Z4
π
Solution. 73.18
1
Z1 Z2 Z1
1
x − 1
x − 1 1 x − = 12 − x when x ≤ 1
dx = dx + x − dx 2
1
2
2 2 2 x −
2 = x − 21 when x ≥ 1
2
0 0 1
2
1
Z 2 Z1
1 1
= − x dx + x− dx
2 2
0 1
2
2 1 2 1
x x 2 x x
= − + + −
2 2 0 2 2 1
2
1 1 1 1 1 1
= − + + − − −
8 4 2 2 8 4
1
=
4
Z4
√ 2
2. x+ x dx . 10
answer: 533
53
9.4 Fundamental Theorem of Calculus Part I
Problem 75. Differentiate f (x) using the Fundamental Theorem of Calculus part 1.
Zx Zx
2
2
1. f (x) = sin t dt 3. f (x) = t2 dt. answer: f 0 (x) = 2x5
1 0
x
answer: sin x
2
Ze
4. f (x) = t3 dt.
ln x
x
Z1 answer: f 0 (x) = e4x −
(ln x)3
2. f (x) = (2 + t4 )5 dt Z x3
x 5. g(x) = cos2 t dt
0
answer: − 2 + x x cos answer: 3x
4 5 3 2 2
d
Rx
Solution. 75.2 We recall that the Fundamental Theorem of Calculus part 1 states that dx a
h(t)dt = h(x) where a is a constant.
We can rewrite the integral so it has x as the upper limit:
Z 1 Z x
f (x) = (2 + 14 )5 dt = − (2 + 14 )5 dt .
x 1
Therefore Z x Z x
d d FTC part 1
− (2 + t4 )5 dt = − (2 + t4 )5 dt = −(2 + x4 )5 .
dx 1 dx 1
Solution. 75.4
x
Ze
d
f 0 (x) = t3 dt
dx
ln x0 x
Z Ze
d
= t3 dt + t3 dt
dx
ln x ln x 0
x
Z Ze
d
= − t3 dt + t3 dt .
dx
0 0
Z u
d
The Fundamental Theorem of Calculus part I states that for an arbitrary constant a, g(t)dt = g(u) (for a continuous g). We
du a
use this two compute the two derivatives:
ln x u
Z Z
d d
t3 dt = t3 dt Set u = ln x
dx dx
0 0
du
= u3 ·
dx
(ln x)3
=
ex x
Z Zw
d d
t3 dt = t3 dt Set w = ex
dx dx
0 0
dw 3
= w ·
dx
= e3x ex = e4x
(ln x)3
f 0 (x) = e4x − .
x
54
9.5 Integration with The Substitution Rule
9.5.1 Substitution in Indefinite Integrals
Problem 76. Evaluate the indefinite integral. The answer key has not been proofread, use with caution.
Z Z Z √
1. (1 + 3x)9 dx. 10. x(2x + 5)2014 dx. sin t
19. √ dt.
30
t
1612 8064
+ C. answer: +C (2x + 5) − (2x + 5) answer:
(1 + 3x)10 2015 1 2016 1 t +C answer: −2 cos
√
√
Z Z p
x3
Z
2. 2x + 1 dx. 11. x2 + 1 dx.
20. sec2 t tan3 tdt.
3 5 3
+C answer: x +1 2 − x +1 2 +C answer:
(2x + 1) 2 1 2 5 1 2 3
4
3 +C answer:
tan4 t
√
Z Z
3
3. (3x + 2) 2.4
dx. 12. x sin 2 + x 2 dx. Z
10.2 3 21. cos4 t sin tdt.
+C answer: +C 2 + x2 cos answer: −
(3x + 2)3.4 3 2
! 5
answer: −
cos πx
Z p Z cos5 t
7. 2 3
(x + 1)(x + 3x) dx. 5 16. dt.
sin t Z
answer: ln | sin t| + C
sec3 t tan tdt.
18
+C
x3 +3x
answer:
25.
6 Z
Z
x2 17. tan tdt. +C
3
answer:
8. √
3
dx.
2
1+x
3 3
+C answer:
sec3 t
1 + x3 1 2 answer: − ln | cos t| + C
Z
√
Z Z
t sin t2 dt.
x2 26.
9. 1 + x dx. 18. cot(2t)dt.
5 3 7 2 2
(x − 1) 2 − (x − 1) 2 + (x − 1) 2 + C answer: ln | sin(2t)| + C answer: cos t +C answer: −
2
7 4 5 2 3 2 1 1
Solution. 76.1 We present two solution variants. The variants are equivalent. The only difference between them is that they use two
interchangeable notations for differentials. Both variants are acceptable both when taking tests and writing scientific texts.
Variant I
Z Z Set
9 9 d(3x) u = 1 + 3x
(1 + 3x) dx = (1 + 3x)
3 du = 3dx
Z dx = 13 du
du
= u9
Z 3
1
= u9 du
3 10
= 30 u + C = (1+3x)
1 10
30 + C.
55
Variant II This variant is equivalent to the previous but uses the differential notation.
Z Z
9 d(3x)
(1 + 3x) dx = (1 + 3x)9 differentials are linear: d(3x) = (3x)0 dx = 3dx
Z 3
d(1 + 3x)
= (1 + 3x)9 differentials don’t change when we add constants
Z 3
1
= u9 du Set u = 1 + 3x
3 10
= 30 u + C = (1+3x)
1 10
30 + C.
Problem 77. Evaluate the integral. The answer key has not been proofread, use with caution.
Z Z Z
dx
1. . 7. esin t cos tdt. 13. cot xdx.
3x + 5
3 answer: ln | sin x| + C
answer: 1 ln |3x + 5| + C answer: esin t + C
Z Z x
dx Z
14. cot dx
2. . 8. e cot x 2
csc xdx. 2
2 − 3x 2
3
answer: − 1 ln |2 − 3x| + C x + C answer: 2 ln sin
answer: −ecot x + C
Z Z
tan(2x)dx.
x x
Z
3. e cos(e )dx. x 15.
9. dx.
answer: sin ex + C
1 + x2 2
answer: − 1 ln | cos(2x)| + C
2
answer: 1 ln(x2 + 1) + C
x4 + 3x
Z
(ln x)3
Z
4. dx.
Z
cos (ln x) 16. dx
x 10. dx. x2
4
(ln x)4
+C answer: x + 3 ln |x| + C
3
answer:
x3
answer: sin(ln x) + C
√
Z Z
ex
3
Z
5. ex + 1 dx sin(2x) 17. x2 ex dx
11. dx.
3 2 + cos2 x 3
(e + 1) 2 + C answer: e +C answer:
x
3 2 answer: − ln(2 + cos2 x) + C 1 x3
√
Z Z
Z
cos x arctan x
6. ex 1 − ex dx. 12. dx 18. dx.
sin x 1 + x2
3 2
1 − ex 2 + C answer: − 2 +C answer:
3 answer: ln | sin x| + C (arctan x)2
Solution. 77.5.
√ √
Z Z
x
e ex + 1 dx = ex + 1 d (ex )
√
Z
= ex + 1 d (ex + 1) Set u = ex + 1
√
Z
= u du
2 3
= u2 + C
3
2 x 3
= (e + 1) 2 + C
3
56
Solution. 77.11 Z Z
sin(2x) 2 cos x sin xdx
dx = use sin(2x) = 2 sin x cos x
2 + cos2 x Z 2 + cos2 x
2 cos xd(− cos x)
= 2
use d(cos x) = − sin xdx
Z 2 + cos x
2ud(u)
= − 2
set u = cos x
Z 2 + u 2
d 2+u
= − use d(u2 + 2) = 2udu
Z 2 + u2
dz
= − set z = 2 + u2
z
= − ln |z| + C Substitute back z = u2 + 2
u2 + 2 is positive
= − ln(u2 + 2) + C ⇒ omit the abs. value
Substitute back u = cos x
= − ln(cos2 x + 2) + C.
Z1
2
2. xe−x dx.
0
2
answer: 1−e
−1
Z1
ex + 1
3. dx.
ex + x
0
answer: ln(e + 1)
Z2
x
4. dx.
2x2 +1
1
4
answer: 1 ln 3
1
Z4
x
5. √ dx.
1 − 3x2
0
16 3
13 1− answer: 1
q
Solution. 78.4
Z2 x=2
1 2
x=2
4 d(2x ) d(2x2 + 1)
Z Z
x 1
2
dx = = Set u = 2x2 + 1
2x + 1 2x2 + 1 4 2x2 + 1
1 x=1 x=1
x=2
u=9
Z
1 du 1 9 1 ln 3
= = [ln u]3 = (ln 9 − ln 3) =
4 u 4 4 4
x=1
u=3
57
10 First Applications of Integration
10.1 Area Between Curves
Problem 79. 1. Find the area of the region bounded by the curves y = 2x2 and y = 4 + x2 .
3
answer: 32
Solution. 79.2. x = 4 − y 2 is a parabola (here we consider x as a function of y). y = −x + 2 implies that x = 2 − y and so the two
curves intersect when
4 − y2 = 2 − y
2
−y + y + 2 = 0
−(y + 1)(y − 2) = 0
y = −1 or 2 .
As x = 2 − y, this implies that x = 0 when y = 2 and x = 3 when y = −1, or in other words the points of intersection are (0, 2) and
(3, −1). Therefore we the region is the one plotted below. Integrating with respect to y, we get that the area is
Z2 Z2
4 − x2 − (−x + 2) dy = −y 2 + y + 2 dy
A =
−1 −1
3 2
y2 (−1)3 (−1)2
y 8
= − + + 2y =− +2+4− − + −2
3 2 −1 3 3 2
9
= .
2
y = −x + 2
x = 4 − y2
Problem 80.
1. Consider the region bounded by the curves y = 2x2 − x + 1 and y = x2 + 1. What is the volume of the solid obtained by rotating
this region about the line x = 0?
5
answer: 2 π.
2. Consider the region bounded by the curves y = 1 − x2 and y = 0. What is the volume of the solid obtained by rotating this
region about the line y = 0?
15
answer: 16π
3. Consider the region bounded by the curves y = x2 and x = y 2 . What is the volume of the solid obtained by rotating this region
about the line x = 2?
30
answer: 31π
58
Solution. 80.1 First, plot y = 2x2 − x + 1 and y = x2 + 1.
y
router
rinner
router x
rinner
z x
2. Consider the region bounded by the vurves y = x2 and y = 2 − x2 . Use the method of cylindrical shells to find the volume of the
solid obtained by rotating this region about the line x = 1. 3
answer: 16 π
59