Question

Find the average rate of change of $g\left(x\right)=x^3-2x+1$

(a) From $-3$to$-2$

(b) From$-1$to$1$

(c) From$1$to$3$

Short answer

(a) The average rate of change is $17.$

(b) The average rate of change is $-1.$

(c) The average rate of change is$11.$

Step-by-step solution

Step 1. Given information.

We have:

$g\left(x\right)=x^3-2x+1$

Part a. Step 1. Find the average rate of change from -3 to -2.

The function is $g\left(x\right)=x^3-2x+1$.

Average rate of change is given by:

localid="1646219921983" $\frac{∆y}{∆x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, $a=-3,b=-2$:

localid="1646220011891" $$\begin{gathered} \frac{∆y}{∆x}=\frac{f(-3)-f(-2)}{-3-\left(-2\right)} \\ =\frac{{\left(-3\right)}^3-2(-3)+1-\left({\left(-2\right)}^3-2(-2)+1\right)}{-3+2} \\ =17 \end{gathered}$$

Part b. Step 1. Find the average rate of change from -1 to 1.

The function is $g\left(x\right)=x^3-2x+1$.

Average rate of change is given by:

$\frac{∆y}{∆x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, $a=-1,b=1$

role="math" localid="1646219347984" $$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(1\right)-f(-1)}{-1-1} \\ =\frac{{\left(1\right)}^3-2\left(1\right)+1-\left({\left(-1\right)}^3-2(-1)+1\right)}{-3+2} \\ =-1 \end{gathered}$$

Part c. Step 1. Find the average rate of change from 1 to 3.

The function is $g\left(x\right)=x^3-2x+1$.

Average rate of change is given by:

$\frac{∆y}{∆x}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here, $a=1,b=3$:

$$\begin{gathered} \frac{∆y}{∆x}=\frac{f\left(3\right)-f\left(1\right)}{3-1} \\ =\frac{{\left(3\right)}^3-2\left(3\right)+1-\left({\left(1\right)}^3-2\left(1\right)+1\right)}{2} \\ =11 \end{gathered}$$