Salmabad, Kingdom of Bahrain

_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024

ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

1. Evaluate the function g(x) = x − 3x + 2 for x = a – 2.

2

g(x) = x^2 - 3x + 2
g(a - 2) = (a - 2)^2 - 3(a - 2) + 2
g(a - 2) = (a^2 - 4a + 4) - (3a - 6) + 2
= a^2 - 4a + 4 - 3a + 6 + 2
= a^2 - 7a + 12

2.

a. f(0):
f(0) = (0)² + 3(0) – 4
f(0) = 0 + 0 – 4
f(0) = -4

b. f(2):
f(2) = (2)² + 3(2) – 4
f(2) = 4 + 6 – 4
f(2) = 6

c. f(h):
f(h) = (h)² + 3(h) – 4
f(h) = h² + 3h - 4

d. f(2h):
f(2h) = (2h)² + 3(2h) – 4
f(2h) = 4h² + 6h – 4
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

e. f(2x):
f(2x) = (2x)² + 3(2x) – 4
f(2x) = 4x² + 6x - 4

f. f(x + h):
f(x + h) = (x + h)² + 3(x + h) – 4
f(x + h) = x² + 2xh + h² + 3x + 3h – 4

g. f(x) + f(h):
f(x) + f(h) = (x)² + 3(x) - 4 + (h)² + 3(h) – 4
f(x) + f(h) = x² + 3x - 4 + h² + 3h – 4

3. If f(x) = 3x − 1 and g(x) = x2, then what is ( f ∘ g ) ( x ) ?

(f∘g)(x) = f(g(x))
(f∘g)(x) = f(x²)
(f∘g)(x) = 3(x²) - 1
(f∘g)(x) = 3x² - 1

4.

(2(1/2) - 1) / (4(1/2)² - 1)
= (1 - 1) / (1 - 1)
=0/0
(2x - 1) / [(2x + 1)(2x - 1)]
(2x - 1) / [(2x + 1)(2x - 1)] = 1 / (2x + 1)
1 / (2(1/2) + 1)
= 1 / (1 + 1)
=1/2
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

5.
Dividing the numerator and denominator by x³
(5x²/x³ - 7x/x³)/(4x³/x³ + 1/x³)
(5/x - 7/x²)/(4 + 1/x³)
(0 - 0)/(4 + 0)
the limit as x approaches infinity for the expression (5x²-7x)/(4x³+1) is 0.

6.

7.

the function F(x) = 1/x - sqrt((x + 6) / (x² + 1)) + (3x² + 5) + sin(x) is continuous for all real values of x.

8.
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

9. Find the equation of the line through (1,1) and (−5,−3) in the form
y = mx + b.
m = (y2 - y1) / (x2 - x1)
m = (-3 - 1) / (-5 - 1)
m = -4 / -6
m = 2/3
1 = (2/3)(1) + b
1 = 2/3 + b
b = 1 - 2/3
b = 1/3
y = (2/3)x + 1/3
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

10.

d = √((x2 - x1)² + (y2 - y1)²)

AB:
dAB = √((10 - 1)² + (5 - 3)²)
= √(9² + 2²)
= √(81 + 4)
= √85
BC:
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

dBC = √((2 - 10)² + (1 - 5)²)

= √((-8)² + (-4)²)
= √(64 + 16)
= √80
AC:
dAC = √((2 - 1)² + (1 - 3)²)
= √(1² + (-2)²)
= √(1 + 4)
= √5
AC² + BC² = AB²
(√5)² + (√80)² = (√85)²
5 + 80 = 85
85 = 85
Since the equation is true, we can conclude that the triangle with vertices A(1, 3), B(10, 5), and C(2,
1) is a right angle triangle.

11.
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

12.

A)

Slope of AB = (-2 - (-6)) / (3 - 4) = 4 / -1 = -4

Slope of BC = (2 - (-2)) / (5 - 3) = 4 / 2 = 2

Slope of AC = (2 - (-6)) / (5 - 4) = 8 / 1 = 8

B)

Tangent of angle A = Tangent of angle opposite to side BC = Tangent of angle opposite to slope of BC = Tangent of
arctan(2) = approximately 1.107

Tangent of angle B = Tangent of angle opposite to side AC =

Tangent of angle opposite to slope of AC = Tangent of
arctan(8) = approximately 6.799

Tangent of angle C = Tangent of angle opposite to side AB =

Tangent of angle opposite to slope of AB = Tangent of
arctan(-4) = approximately -1.157

13.
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

For y = 2x - 8, the slope is 2.

For y = x, the slope is 1.

tan(θ) = (m1 - m2) / (1 + m1*m2)

In this case, m1 = 2 and m2 = 1.

tan(θ) = (2 - 1) / (1 + 2*1)

tan(θ) = 1 / 3

14.

Line 1: 3x - 5y + 19 = 0

y = (3/5)x + 19/5

The slope is 3/5.

2: 10x + 6y - 50 = 0

y = -(10/6)x + 25/3

The slope is -10/6, which simplifies to -5/3.

(3/5) * (-5/3) = -1

3x - 5y + 19 = 0

10x + 6y - 50 = 0

18x - 30y + 114 = 0

50x + 30y - 250 = 0

68x - 136 = 0

x=2
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

3(2) - 5y + 19 = 0

6 - 5y + 19 = 0

-5y + 25 = 0

-5y = -25

y=5

the lines intersect at the point (2, 5).

15.

A.

1. f(x) = (3x⁴ - 2x² + 8)³

f'(x) = 3(3x⁴ - 2x² + 8)²(12x³ - 4x)
2. f(x) = (x⁴ - 2)³(x⁵ + 4x)²
f'(x) = 3(x⁴ - 2)²(4x³)(x⁵ + 4x)² + (x⁴ - 2)³(2)(5x⁴ + 4)
3. y = ³√((4x + 3)/(x - 5))
y' = (1/3)(4x + 3)^(-2/3) * ((x - 5)(1) - (4x + 3)(1))/((x - 5)²)
= (1/3)(4x + 3)^(-2/3) * (x - 5 - 4x - 3)/((x - 5)²)
= -(1/3)(4x + 3)^(-2/3) * (3x +8)/((x - 5)²)

B.
1. f(x) = sin(2x + 5)
f'(x) = cos(2x + 5) * 2
 Salmabad, Kingdom of Bahrain
_ X __1st Tri, __ _2nd Tri, __ _1st Tri, SY 2023-2024
ASSIGMENT 1
MATH 406 -631 -632 –DIFFERENTIAL CALCULUS AND ANALYTIC GEOMETRY

2. f(x) = sec(√x)
f'(x) = sec(√x) * tan(√x) * (1/2√x)

3. J(x) = cos(x) * sin(3x)

J'(x) = -sin(x) * sin(3x) + cos(x) * 3cos(3x)

4. f(θ) = tan(sin(2θ))
f'(θ) = sec²(sin(2θ)) * cos(2θ) * 2

C.
1. xy² + 4y³ = x - 2y
d/dx(xy² + 4y³) = d/dx(x - 2y)
y² + 2xy(dy/dx) + 12y²(dy/dx) = 1 - 2(dy/dx)
(2xy + 12y² + 2)dy/dx = 1 - y²
dy/dx = (1 - y²) / (2xy + 12y² + 2)

2. sin(x+y) + sin(x - y) = 1
d/dx(sin(x+y) + sin(x - y)) = d/dx(1)
cos(x+y)(1+dy/dx) + cos(x - y)(1 - dy/dx) = 0
cos(x+y) + cos(x - y) + cos(x+y)dy/dx - cos(x - y)dy/dx = 0
(2cos(x+y) - 2cos(x - y))dy/dx = -cos(x+y) - cos(x - y)
dy/dx = (-cos(x+y) - cos(x - y)) / (2cos(x+y) - 2cos(x - y))