## Calculus (3rd Edition)

$f(x)$ increasing on $\left( -\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right )$ and decreasing on $\left(-\infty, -\frac{1+\sqrt{5}}{2}\right )\cup\left( \frac{1+\sqrt{5}}{2},\infty\right )$ $f(x)$ has a local maximum at $x= \frac{1+\sqrt{5}}{2}$ and a local minimum at $x=- \frac{1+\sqrt{5}}{2}$.
Given $$y=\frac{2 x+1}{x^{2}+1}$$ Since \begin{align*} f'(x) &=\frac{\frac{d}{dx}\left(2x+1\right)\left(x^2+1\right)-\frac{d}{dx}\left(x^2+1\right)\left(2x+1\right)}{\left(x^2+1\right)^2}\\ &= \frac{-2x^2-2x+2}{\left(x^2+1\right)^2} \end{align*} Then $f'(x)=0$ for $-2x^2-2x+2= 0$. Hence $$\quad x=-\frac{1+\sqrt{5}}{2},\:x=\frac{\sqrt{5}-1}{2}$$ Choose $x=-2,\ \ x=0$ and $x=1$: \begin{align*} f'(-2)&< 0\\ f'(0)&>0\\ f'(1)&<0 \end{align*} Then $f(x)$ is increasing on $\left( -\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right )$ and decreasing on $\left(-\infty, -\frac{1+\sqrt{5}}{2}\right )\cup\left( \frac{1+\sqrt{5}}{2},\infty\right )$. Hence, $f(x)$ has a local maximum at $x= \frac{1+\sqrt{5}}{2}$ and a local minimum at $x=- \frac{1+\sqrt{5}}{2}$.