determine the location and value of the absolute extreme values of f on the given interval, if they exist…

Explanation:

Step1: Find the derivative using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x$, $u^\prime=1$, $v=(x^{2}+7)^{4}$, and $v^\prime = 4(x^{2}+7)^{3}\cdot2x=8x(x^{2}+7)^{3}$. So, $f^\prime(x)=\frac{(x^{2}+7)^{4}-x\cdot8x(x^{2}+7)^{3}}{(x^{2}+7)^{8}}=\frac{(x^{2}+7)^{3}[(x^{2}+7)-8x^{2}]}{(x^{2}+7)^{8}}=\frac{7 - 7x^{2}}{(x^{2}+7)^{5}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$. Then $\frac{7 - 7x^{2}}{(x^{2}+7)^{5}}=0$. Since the denominator $(x^{2}+7)^{5}>0$ for all real $x$, we solve $7 - 7x^{2}=0$. $7 - 7x^{2}=0$ implies $x^{2}=1$, so $x=\pm1$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x=-2,x = - 1,x = 1,x = 2$. $f(-2)=\frac{-2}{((-2)^{2}+7)^{4}}=\frac{-2}{(4 + 7)^{4}}=\frac{-2}{11^{4}}$. $f(-1)=\frac{-1}{((-1)^{2}+7)^{4}}=\frac{-1}{8^{4}}$. $f(1)=\frac{1}{(1^{2}+7)^{4}}=\frac{1}{8^{4}}$. $f(2)=\frac{2}{(2^{2}+7)^{4}}=\frac{2}{11^{4}}$.

Answer:

A. The absolute maximum value is $\frac{1}{4096}$ at $x = 1$.