3. -/1 points details my notes tanapcalc10 4.4.014. find the absolute maximum value and the absolute minimum…

Explanation:

Step1: Find the derivative

Use the quotient - rule. If $f(x)=\frac{u}{v}$ where $u = x$ and $v=4 + x^{2}$, then $f^\prime(x)=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u^\prime=1$ and $v^\prime = 2x$. So $f^\prime(x)=\frac{4 + x^{2}-x(2x)}{(4 + x^{2})^{2}}=\frac{4 - x^{2}}{(4 + x^{2})^{2}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$. Then $\frac{4 - x^{2}}{(4 + x^{2})^{2}}=0$. Since $(4 + x^{2})^{2}>0$ for all real $x$, we solve $4 - x^{2}=0$. Factoring gives $(2 - x)(2 + x)=0$, so $x=- 2$ or $x = 2$.

Step3: Evaluate the function at critical points and endpoints (in this case, since the domain is all real numbers, we just use the critical points)

$f(-2)=\frac{-2}{4+( - 2)^{2}}=-\frac{2}{8}=-\frac{1}{4}$. $f(2)=\frac{2}{4 + 2^{2}}=\frac{2}{8}=\frac{1}{4}$. As $x\to\pm\infty$, $\lim_{x\to\pm\infty}\frac{x}{4 + x^{2}}=\lim_{x\to\pm\infty}\frac{\frac{x}{x^{2}}}{\frac{4}{x^{2}}+1}=\lim_{x\to\pm\infty}\frac{\frac{1}{x}}{\frac{4}{x^{2}}+1}=0$.

Answer:

maximum: $\frac{1}{4}$ minimum: $-\frac{1}{4}$