Chapter 10.2, Problem 41E

To determine

To find: The cube root function that passes through the point $\left(3,4\right)$ and has the average rate of change of $-1$ over the interval of $x=-5$ to $x=2$ .

Answer to Problem 41E

The required function is $f\left(x\right)=-7{(x-3)}^{1/3}+4$

Explanation of Solution

Given information: Function passes through $\left(3,4\right)$ and has an average rate of change of $-1$ over the interval $x=-5tox=2.$

Calculation:

Consider the simplest cubic function that passes through the point $\left(3,4\right)$ , i.e.,

  $f\left(x\right)=a{(x-3)}^{1/3}+4$ .

Then, the average value of the function is,

  $\begin{array}{c}\frac{f\left(2\right)-f\left(-5\right)}{2-\left(-5\right)}=\frac{\left(a{\left(2-3\right)}^{1/3}\right)+4-\left(a{\left(-5-3\right)}^{1/3}\right)-4}{7}\\ =\frac{a{\left(-1\right)}^{1/3}-a{\left(-8\right)}^{1/3}}{7}\\ =\frac{-a+2a}{7}\\ =\frac{a}{7}\end{array}$

Thus, the average value should be $-1$ that is,

  $\begin{array}{l}\frac{a}{7}=-1\\ a=-7\end{array}$

Then, the required function is,

  $f\left(x\right)=-7{(x-3)}^{1/3}+4$