DIFFERENTIATION & FUNCTIONS (Q 6, 7 & 8, PAPER 1)

LESSON NO. 15: INTERSECTING GRAPHS

2005
1
8 Let f ( x) = , x ∈ R , x ≠ 1.
x −1

(i) Find f (−3), f (−1.5), f (0.5), f (1.5), f (5).

(ii) Draw the graph of the function f from x = −3 to x = 5.

(iii) On the same diagram, draw the graph of the function

g(x) = x +1
in the domain −2 ≤ x ≤ 2, x ∈ R.

(iv) Use your graphs to estimate the values of x for which f (x) = g(x).

(v) Find, using algebra, the values of x for which f (x) = g(x).
SOLUTION
8 (i)
1
f ( x) =
x −1
1 1
f (−3) = = − = −0.25
(−3) − 1 4
1 1
f (−1.5) = =− = −0..4
(−1.5) − 1 2.5
1 1
f (0.5) = =− = −2
(0.5) − 1 0.5
1 1
f (1.5) = = =2
(1.5) − 1 0.5
1 1
f (5) = = = 0.25
(5) − 1 4

8 (ii) STEPS
1. Find the gap first by putting the bottom of the function equal to
zero and solving for x.
2. Find other values of f (x) by putting in values of x as given in the
domain.
3. Plot these points by joining up smoothly and continuing towards
the vertical line but never touching it.

1. Put x − 1 = 0 ⇒ x = 1 is the asymptote.

2. You generated points in part (i) which you can use to draw the graph.
Points: (−3, − 0.25), (−1.5, − 0.4), (−2, 0.5), (1.5, 2), (5, 0.25)
3. Draw the graph.
CONT....
8 (iii)
g(x) = x + 1 is a straight line graph so you just need 2 points to draw the graph. Use the end
values of the domain.
x = −2 : g ( x) = x + 1 ⇒ g (−2) = (−2) + 1 = −1 ⇒ (−2, − 1) is a point.
x = 2 : g ( x) = x + 1 ⇒ g (2) = (2) + 1 = 3 ⇒ (2, 3) is a point.
Plot these two points using the same axes and draw a straight line through them.

8 (iv)
Find out where the two graphs intersect and read off the x values.
You can see that x = 1.4 and x = −1.4.

8 (v)
1
f ( x) = g ( x) ⇒ = x + 1 [Multiply across by ( x −1). ]
x −1
⇒ 1 = ( x + 1)( x − 1) [Multiply out the brackets.]
⇒ 1 = x2 − 1
⇒ 2 = x2
⇒x=± 2

f (x)
3

1
-1.4

-3 -2 -1 1 2 3 4 5 x
1.4
-1

-2
x=1

2000
8 (b) (i) Draw the graph of
g ( x) = 1x for −3 ≤ x ≤ 3, x ∈ R and x ≠ 0.
(ii) Using the same axes and the same scales, draw the graph of
h(x) = x + 1 for −3 ≤ x ≤ 3, x ∈ R.
(iii) Use your graphs to estimate the values of x for which
1
x = x + 1.
CONT....
SOLUTION
8 (b) (i)
DRAWING RECIPROCAL GRAPHS
STEPS
1. Find the gap first by putting the bottom of the function equal to
zero and solving for x.
2. Find other values of f (x) by putting in values of x as given in the
domain.
3. Plot these points by joining up smoothly and continuing towards
the vertical line but never touching it.

1. Put x = 0. The line x = 0 (i.e. the y-axis) represents the gap or asymptote.
1 1
2. x = −3 : f (−3) = = − = −0.33 ⇒ (−3, − 0.33)
−3 3
1 1
x = −2 : f (−2) = = − = −0.5 ⇒ (−2, − 0.5)
−2 2
1
x = −1 : f (−1) = = −1 ⇒ (−1, − 1)
−1
x = 0 : This is the gap where the function is not defined.
1
x = 1 : f (1) = = 1 ⇒ (1, 1)
1
1
x = 2 : f (2) = = 0.5 ⇒ (2, 0.5)
2
1
x = 3 : f (3) = = 0.33 ⇒ (3, 0.33)
3 f (x)
3. Plot the reciprocal graph. 4

-1.6 1

-3 -2 -1 1 2 3
-1
0.6
-2

-3
x=0

8 (b) (ii)
As these graphs are straight lines, you only need to plot the
first and last points in the domain.
h( x ) = x + 1
⇒ h(−3) = −3 + 1 = −2 ⇒ (−3, − 2) is a point on the straight line.
⇒ h(3) = 3 + 1 = 4 ⇒ (3, 4) is a point on the straight line.

8 (b) (iii)
Read off the x values of where the graphs intersect.
∴ x = −1.6, 0.6
1999
8 Let f ( x) = 2 x3 − 5 x 2 − 4 x + 3 for x ∈ R.
(i) Complete the table

x −1.5 −1 0 1 2 3 3.5
f (x) −9 13.5

(ii) Find the derivative of f (x).

Calculate the co-ordinates of the local minimum and show that the co-ordinates
of the local maximum are (− 13 , 100
27 ).
(iii) Draw the graph of
f ( x) = 2 x3 − 5 x 2 − 4 x + 3
for −1.5 ≤ x ≤ 3.5.
(iv) Write the equation 2 x3 − 5 x 2 − 6 x + 6 = 0 in the form
2 x3 − 5 x 2 − 4 x + 3 = ax + b, a, b ∈ Z.
Hence, use your graph to estimate the solutions of the equation
2 x3 − 5 x 2 − 6 x + 6 = 0.
SOLUTION
8 (i)
f ( x) = 2 x3 − 5 x 2 − 4 x + 3
⇒ f (−1.5) = 2(−1.5)3 − 5(−1.5) 2 − 4(−1.5) + 3 = −9 [Use your calculator.]
Find the other values of x in the same way.

x −1.5 −1 0 1 2 3 3.5
f (x) −9 0 3 −4 −9 0 13.5

8 (ii) STEPS FOR FINDING THE LOCAL MAXIMUM AND LOCAL MINIMUM OF A FUNCTION:
STEPS
dy d2y
1. Differentiate the function to find . Differentiate again to find .
dx dx 2
dy
2. Set = 0 and solve for x to find the turning points.
dx

d2y
3. Substitute the turning points into to decide if they are a local
dx 2
maximum or a local minimum.
4. Find the y coordinates of the turning points by substituting the x values
back into the equation of the original function.

CONT....
1. y = f ( x) = 2 x3 − 5 x 2 − 4 x + 3
dy
= f ′( x) = 6 x 2 − 10 x − 4
dx
d2y
= f ′′( x) = 12 x − 10
dx 2
dy
2. = 0 ⇒ 6 x 2 − 10 x − 4 = 0
dx
⇒ 3x 2 − 5 x − 2 = 0
⇒ (3 x + 1)( x − 2) = 0
∴ x = − 13 , 2
⎛ d2y ⎞
⎛ d2y ⎞ Local Maximum: ⎜ 2 ⎟ < 0
3. ⎜ 2 ⎟ = 12(− 13 ) − 10 = −4 − 10 = −14 < 0 ⎝ dx ⎠ TP
⎝ dx ⎠ x =− 1 3 ....... 7
⎛ d2y ⎞
⎛d y⎞
2
Local Minimum: ⎜ 2 ⎟ > 0
⎜ 2 ⎟ = 12(2) − 10 = 24 − 10 = 14 > 0 ⎝ dx ⎠ TP
⎝ dx ⎠ x = 2
27 ⇒ ( − 3 ,
4. x = − 13 : y = 2(− 13 )3 − 5(− 13 ) 2 − 4(− 13 ) + 3 = 100 1 100
27 ) is a local maximum.
x = 2 : y = 2(2)3 − 5(2) 2 − 4(2) + 3 = −9 ⇒ (2, − 9) is a local minimum.

8 (iii) f (x) 14
Draw the cubic graph using the information 12
from the previous parts. 10
Points: (−1.5, − 9), (−1, 0), (0, 3), (1, − 4) 8
(2, − 9), (3, 0), (3.5, 13.5) 6

Local maximum: (− 13 , 100

27 ) = (−0.33, 3.7) 4

2
Local minimum: (2, − 9) x
-2 -1 1 2 3 4
-2
8 (iv) -4
2 x3 − 5 x 2 − 6 x + 6 = 0 -6
⇒ 2 x − 5x − 4 x + 3 = 2 x − 3
3 2
-8

-10
Let h( x) = 2 x − 3
∴ f ( x) = h( x). f (x) 14

12
h(x) is a straight line. You want to find where
10
the straight line and the cubic graph intersect.
Graph h(x) by using the first and last points of 8

the domain. 6

h(−1.5) = 2(−1.5) − 3 = −3 − 3 = −6 -1.4

4
0.7
⇒ (−1.5, − 6) is a point on the graph. 2
x
h(3.5) = 2(3.5) − 3 = 7 − 3 = 4 -2 -1
-2
1 2 3 4

⇒ (3.5, 4) is a point on the graph. -4

3.2

-6
You can see the x values of the places where the
graphs intersect: -8

∴ x = −1.4, 0.7, 3.2 -10

1997
8 (c) Draw a graph of
1
g ( x) =
x+2
for 0 ≤ x ≤ 4, x ∈ R.
Using the same axes and the same scales draw the graph of
h( x) = x − 2.
Show how your graphs may be used to estimate the value of 5.
SOLUTION DRAWING RECIPROCAL GRAPHS
STEPS
1. Find the gap first by putting the bottom of the function equal to
zero and solving for x.
2. Find other values of f (x) by putting in values of x as given in the
domain.
3. Plot these points by joining up smoothly and continuing towards
the vertical line but never touching it.

1. Put x + 2 = 0 ⇒ x = −2. This line represents the gap or asymptote.

1
2. g ( x) =
x+2
1 1
g (0) = = ⇒ (0, 12 ) is a point on the graph.
0+2 2
1 1
g (1) = = ⇒ (0, 13 ) is a point on the graph.
1+ 2 3
1 1
g (2) = = ⇒ (0, 14 ) is a point on the graph.
2+2 4
1 1
g (3) = = ⇒ (0, 15 ) is a point on the graph.
3+ 2 5
1 1
g (4) = = ⇒ (0, 16 ) is a point on the graph.
4+2 6
3. Plot the graph. g(x)
3

-3 -2 -1 1 2 3 4 5 x

-1

-2
x = -2
CONT....
The graph of h( x) = x − 2 is a straight line graph. You just need to get 2 points on it to graph
it. Choose the end points of the domain.
x = 0 : h(0) = (0) − 2 = −2 ⇒ (0, − 2) is a point on the graph.
x = 4 : h(4) = (4) − 2 = 2 ⇒ (4, 2) is a point on the graph.

g(x) and h(x)

3

-3 -2 -1 1 2 3 4 5 x
2.2
-1

-2
x = -2

STEPS
1. Draw each graph on the same diagram using the same scales.
2. Mark the points of intersection and read off their x value.
3. Solve f (x) = g(x) exactly by algebra.
4. Hence, estimate the solution.

1. Graphs drawn as above.

2. x = 2.2 is their point of intersection.
1
3. g ( x) = h( x) ⇒ = x−2
x+2
⇒ 1 = ( x − 2)( x + 2) ⇒ 1 = x 2 − 4 ⇒ x 2 = 5
∴x = 5
4. ∴ 5 ≈ 2.2