Question

find the average rate of change of $g(x)=4x^{4}+\frac{2}{x^{3}}$ on the interval -2, 1.

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = g(x)$ on the interval $[a,b]$ is $\frac{g(b)-g(a)}{b - a}$. Here, $a=-2$, $b = 1$, and $g(x)=4x^{4}+\frac{2}{x^{3}}$.

Step2: Calculate $g(1)$

Substitute $x = 1$ into $g(x)$:
$g(1)=4\times1^{4}+\frac{2}{1^{3}}=4 + 2=6$.

Step3: Calculate $g(-2)$

Substitute $x=-2$ into $g(x)$:
$g(-2)=4\times(-2)^{4}+\frac{2}{(-2)^{3}}=4\times16+\frac{2}{-8}=64-\frac{1}{4}=\frac{256 - 1}{4}=\frac{255}{4}$.

Step4: Calculate the average rate of change

$\frac{g(1)-g(-2)}{1-(-2)}=\frac{6-\frac{255}{4}}{3}=\frac{\frac{24 - 255}{4}}{3}=\frac{24 - 255}{12}=\frac{-231}{12}=-\frac{77}{4}$.

Answer:

$-\frac{77}{4}$