Exercise 101 Page 90

Practice makes perfect

a To sketch the functions, we must calculate a few data points through which the graph passes. We can do this by using a value table.

|c|c|c| [-1em] x & x^3+1 & f(x) [0.2em] [-1em] -2 & ( -2)^3+1 & -7 [0.2em] [-1em] -1 & ( -1)^3+1 & 0 [0.2em] [-1em] 0 & 0^3+1 & 1 [0.2em] [-1em] 1 & 1^3+1 & 2 [0.2em] [-1em] 2 & 2^3+1 & 9 [0.2em] Now we can sketch f(x).

Let's make a second value table for g(x). |c|c|c| [-1em] x & (x+1)^2 & g(x) [0.2em] [-1em] -2 & ( -2+1)^2 & 1 [0.2em] [-1em] -1 & ( -1+1)^2 & 0 [0.2em] [-1em] 0 & ( 0+1)^2 & 1 [0.2em] [-1em] 1 & ( 1+1)^2 & 4 [0.2em] [-1em] 2 & ( 2+1)^2 & 9 [0.2em] Now we can sketch g(x) as well.

b To solve the equation, we have to set substitute f(x) with 9 in our function and solve for x.

f(x)=x^3+1

Substitute

f(x)= 9

9=x^3+1

SubEqn

LHS-1=RHS-1

8=x^3

RearrangeEqn

Rearrange equation

x^3=8

RadicalEqn

sqrt(LHS)=sqrt(RHS)

x=2

c To solve the equation, we have to set substitute g(x) with 0 in our function and solve for x.

g(x)=(x+1)^2

Substitute

g(x)= 0

0=(x+1)^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

0=x+1

SubEqn

LHS-1=RHS-1

-1 =x

RearrangeEqn

Rearrange equation

x= - 1

d Like in Parts B and C, we have to substitute f(x) with -12 and solve for x.

f(x)=x^3+1

Substitute

f(x)= -12

- 12=x^3+1

SubEqn

LHS-1=RHS-1

- 13 = x^3

RearrangeEqn

Rearrange equation

x^3= - 13

RadicalEqn

sqrt(LHS)=sqrt(RHS)

x=sqrt(- 13)

e From Part A, we know that g(x) is a parabola with a minimum value that's on the x-axis. This means it cannot go below x=0 and therefore g(x)=-12 does not have any solutions. By graphing g(x) and y=-12 we see that the graphs never intersect.