find the intervals on which f is increasing and the intervals on which it is decreasing. f(x)=tan^(-1)(x/(x^(…

Explanation:

Step1: Recall the derivative of arctangent

The derivative of $y = \tan^{- 1}(u)$ is $y'=\frac{u'}{1 + u^{2}}$. Let $u=\frac{x}{x^{2}+5}$, then we first find the derivative of $u$ using the quotient - rule. The quotient - rule states that if $u=\frac{v}{w}$, then $u'=\frac{v'w - vw'}{w^{2}}$, where $v = x$, $v'=1$, $w=x^{2}+5$, $w' = 2x$. So $u'=\frac{1\cdot(x^{2}+5)-x\cdot(2x)}{(x^{2}+5)^{2}}=\frac{x^{2}+5 - 2x^{2}}{(x^{2}+5)^{2}}=\frac{5 - x^{2}}{(x^{2}+5)^{2}}$.

Step2: Find the derivative of $f(x)$

By the chain - rule, $f'(x)=\frac{\frac{5 - x^{2}}{(x^{2}+5)^{2}}}{1+\left(\frac{x}{x^{2}+5}\right)^{2}}$. Simplify the denominator: $1+\left(\frac{x}{x^{2}+5}\right)^{2}=\frac{(x^{2}+5)^{2}+x^{2}}{(x^{2}+5)^{2}}$. Then $f'(x)=\frac{5 - x^{2}}{(x^{2}+5)^{2}+x^{2}}=\frac{5 - x^{2}}{x^{4}+10x^{2}+25+x^{2}}=\frac{5 - x^{2}}{x^{4}+11x^{2}+25}$.

Step3: Find the critical points

Set $f'(x) = 0$. Since the denominator $x^{4}+11x^{2}+25>0$ for all real $x$ (because $x^{4}\geq0$, $11x^{2}\geq0$ and $25>0$), we set the numerator equal to $0$. So $5 - x^{2}=0$, which gives $x=\pm\sqrt{5}$.

Step4: Test the intervals

We consider the intervals $(-\infty,-\sqrt{5})$, $(-\sqrt{5},\sqrt{5})$ and $(\sqrt{5},\infty)$. For $x\in(-\infty,-\sqrt{5})$, let $x=-3$. Then $f'(-3)=\frac{5 - 9}{(-3)^{4}+11\times(-3)^{2}+25}=\frac{-4}{81 + 99+25}<0$. For $x\in(-\sqrt{5},\sqrt{5})$, let $x = 0$. Then $f'(0)=\frac{5-0}{0 + 0+25}>0$. For $x\in(\sqrt{5},\infty)$, let $x = 3$. Then $f'(3)=\frac{5 - 9}{3^{4}+11\times3^{2}+25}=\frac{-4}{81 + 99+25}<0$.

Answer:

B. The function is decreasing on the open intervals $(-\infty,-\sqrt{5}),(\sqrt{5},\infty)$ and increasing on the open interval $(-\sqrt{5},\sqrt{5})$