find the absolute maximum and absolute minimum values of $f(x)=\frac{x^{2}-4}{x^{2}+4}$ on the interval…

Explanation:

Step1: Find the derivative of $f(x)$

Use the quotient - rule. If $f(x)=\frac{u(x)}{v(x)}$ where $u(x)=x^{2}-4$ and $v(x)=x^{2}+4$, then $u^\prime(x) = 2x$ and $v^\prime(x)=2x$. The quotient - rule states that $f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v^{2}(x)}$. So, $f^\prime(x)=\frac{2x(x^{2}+4)-2x(x^{2}-4)}{(x^{2}+4)^{2}}=\frac{2x^{3}+8x - 2x^{3}+8x}{(x^{2}+4)^{2}}=\frac{16x}{(x^{2}+4)^{2}}$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. Since $(x^{2}+4)^{2}>0$ for all real $x$, we set the numerator equal to 0. So, $16x = 0$, which gives $x = 0$.

Step3: Evaluate $f(x)$ at the critical point and endpoints

Evaluate $f(x)$ at $x=-5,x = 0,x = 5$. When $x=-5$, $f(-5)=\frac{(-5)^{2}-4}{(-5)^{2}+4}=\frac{25 - 4}{25+4}=\frac{21}{29}$. When $x = 0$, $f(0)=\frac{0^{2}-4}{0^{2}+4}=-1$. When $x = 5$, $f(5)=\frac{5^{2}-4}{5^{2}+4}=\frac{25 - 4}{25+4}=\frac{21}{29}$.

Answer:

a) $\frac{21}{29}$ b) $-1$