Question

how do the average rates of change for the pair of functions compare over the given interval?

$f(x)=3x^{2}$
$g(x)=12x^{2}$
$-6\leq x\leq -4$

the average rate of change of $f(x)$ over $-6\leq x\leq -4$ is $\square$. the average rate of change of $g(x)$ over $-6\leq x\leq -4$ is $\square$. the average rate of change of $g(x)$ is $\square$ times that of $f(x)$.

(simplify your answers. type integers or decimals.)

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\).

Step2: Calculate the average rate of change for \(f(x)=3x^{2}\) over \([-6,-4]\)

Here \(a=-6\), \(b = - 4\).
First, find \(f(-6)\) and \(f(-4)\):
\(f(-6)=3\times(-6)^{2}=3\times36 = 108\)
\(f(-4)=3\times(-4)^{2}=3\times16=48\)
Then, \(\frac{f(-4)-f(-6)}{-4-(-6)}=\frac{48 - 108}{-4 + 6}=\frac{-60}{2}=-30\)

Step3: Calculate the average rate of change for \(g(x)=12x^{2}\) over \([-6,-4]\)

Here \(a=-6\), \(b=-4\)
First, find \(g(-6)\) and \(g(-4)\):
\(g(-6)=12\times(-6)^{2}=12\times36 = 432\)
\(g(-4)=12\times(-4)^{2}=12\times16 = 192\)
Then, \(\frac{g(-4)-g(-6)}{-4-(-6)}=\frac{192-432}{-4 + 6}=\frac{-240}{2}=-120\)

Step4: Find the ratio of the average rate of change of \(g(x)\) to \(f(x)\)

Let \(A_{f}\) be the average rate of change of \(f(x)\) and \(A_{g}\) be the average rate of change of \(g(x)\).
\(\frac{A_{g}}{A_{f}}=\frac{-120}{-30}=4\)

Answer:

The average rate of change of \(f(x)\) over \(-6\leq x\leq - 4\) is \(-30\). The average rate of change of \(g(x)\) over \(-6\leq x\leq - 4\) is \(-120\). The average rate of change of \(g(x)\) is \(4\) times that of \(f(x)\).