Chapter 10, Problem 6CA

To determine

To find: The statements which are true.

Answer to Problem 6CA

The average rate of change of g is $\frac{0.5}{0.05}\approx 10$ times greater than the average rate of f over the interval $x=0$ to $x=8$ .

Explanation of Solution

Given information:

The function: $f\left(x\right)=\frac{1}{2}\sqrt[3]{x+3}$

The graph of the function:

  

The x -intercept of the graph of f is greater than the x -intercept of the graph of g .

The graph of g is always increasing.

The average rate of change of g decreases as x increases.

The average rate of change of f increases as x increases.

The average rate o0f change of g is greater than the average rate of change of f over the interval $x=0$ to $x=8$ .

The given function is $f\left(x\right)=\frac{1}{2}\sqrt[3]{x+3}$ .

From the given graph, it is seen that the x -intercept of the function f is $-3$ when $g$ is 0.

So, the coordinates are $\left(-3,0\right)$ .

So, the graph of both functions f and g are always increasing. Also, the average rate of change for both the functions decreases as x increases.

For the function $g\left(x\right)$ ;

The average rate of change for the points $\left(0,0\right)$ and $\left(8,4\right)$ is,

  $\begin{array}{c}\frac{g\left(8\right)-g\left(0\right)}{8-0}=\frac{4-0}{8-0}\\ =\frac{4}{8}\\ =0.5\end{array}$

For the function $f\left(x\right)=\frac{1}{2}\sqrt[3]{x+3}$ ;

Rewrite the value of the function f when $x=0$ and $x=8$ .

  $\begin{array}{c}f\left(0\right)=\frac{1}{2}\sqrt[3]{0+3}\\ =\frac{1}{2}\sqrt[3]{3}\\ \approx 0.72\end{array}$

And, $\begin{array}{c}f\left(8\right)=\frac{1}{2}\sqrt[3]{8+3}\\ =\frac{1}{2}\sqrt[3]{11}\\ \approx 1.11\end{array}$

The average rate of change is,

  $\begin{array}{c}\frac{f\left(8\right)-f\left(0\right)}{8-0}=\frac{1.11-0.72}{8-0}\\ =\frac{0.39}{8}\\ =0.05\end{array}$

It is seen that the average rate of change of g is $\frac{0.5}{0.05}\approx 10$ times greater than the average rate of f over the interval $x=0$ to $x=8$ .