## Calculus (3rd Edition)

$t = 1$ corresponds to neither a maximum nor a minimum.
Given $$g(t)=\frac{e^{t}}{t^{2}+1}$$ Since \begin{align*} g'(x) &= \frac{\frac{d}{dt}\left(e^t\right)\left(t^2+1\right)-\frac{d}{dt}\left(t^2+1\right)e^t}{\left(t^2+1\right)^2}\\ &= \frac{e^t\left(t^2+1\right)-2te^t}{\left(t^2+1\right)^2} \end{align*} Then $f(x)$ has critical points when \begin{align*} g'(x)&=0\\ \frac{e^t\left(t^2+1\right)-2te^t}{\left(t^2+1\right)^2} &=0\\ e^t[\left(t^2+1\right)-2t]&=0\\ t^2-2t+1&=0 \end{align*} Then $t=1$ is a critical point; to find the interval of increasing and decreasing regions, choose $t=0,\ t=2$ \begin{align*} g'(0)&>0 \\ g'(2)& >0 \end{align*} Hence, $g'(t)$ does not change sign and therefore $t = 1$ corresponds to neither a maximum nor a minimum.