- / 2 points details my notes scalc9 3.1.517.xp.mi. find the absolute maximum and absolute minimum values of…

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, $u = x^{2}-36$, $u^\prime = 2x$, $v=x^{2}+36$, $v^\prime = 2x$. So $f^\prime(x)=\frac{2x(x^{2}+36)-2x(x^{2}-36)}{(x^{2}+36)^{2}}=\frac{2x^{3}+72x - 2x^{3}+72x}{(x^{2}+36)^{2}}=\frac{144x}{(x^{2}+36)^{2}}$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. Since $(x^{2}+36)^{2}>0$ for all real $x$, we solve $\frac{144x}{(x^{2}+36)^{2}}=0$. The numerator must be zero, so $144x = 0$, which gives $x = 0$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x=-36,0,36$. When $x=-36$, $f(-36)=\frac{(-36)^{2}-36}{(-36)^{2}+36}=\frac{1296 - 36}{1296+36}=\frac{1260}{1332}=\frac{35}{37}\approx0.946$. When $x = 0$, $f(0)=\frac{0 - 36}{0+36}=-1$. When $x = 36$, $f(36)=\frac{36^{2}-36}{36^{2}+36}=\frac{1296 - 36}{1296+36}=\frac{1260}{1332}=\frac{35}{37}\approx0.946$.

Answer:

absolute minimum value: $-1$ absolute maximum value: $\frac{35}{37}$