Chapter 2.4, Problem 42E

To determine

To find: The correct choice for the average rate of change of $f\left(x\right)=x^2+x$ over the interval $\left[1,3\right]$.

Answer to Problem 42E

The correct choice is (E) such that the average rate of change of the function $f\left(x\right)=x^2+x$ over the interval $\left[1,3\right]$ is $5$.

Explanation of Solution

Given information: The function is $f\left(x\right)=x^2+x$ and the interval is $\left[1,3\right]$.

The formula for the average rate of change of function $f\left(x\right)$ over the interval $\left[a,b\right]$ is equal to $\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

The average rate of change of function $f\left(x\right)=x^2+x$ over the interval $\left[1,3\right]$ is:

  $\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Substitute $1$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=x^2+x\\ f\left(1\right)=1^2+1\\ f\left(1\right)=1+1\\ f\left(1\right)=2\end{array}$

Now substitute $3$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=x^2+x\\ f\left(3\right)=3^2+3\\ f\left(3\right)=9+3\\ f\left(3\right)=12\end{array}$

Substitute $2$ for $f\left(1\right)$ and $12$ for $f\left(3\right)$ in the formula.

  $\begin{array}{c}\frac{f\left(3\right)-f\left(1\right)}{2}=\frac{12-2}{2}\\ =\frac{10}{2}\\ =5\end{array}$

Therefore, the correct choice is (E) such that the average rate of change of the function $f\left(x\right)=x^2+x$ over the interval $\left[1,3\right]$ is $5$.