= Solution
The <chain rule> and the <quotient rule> give
$$
f'(x)=\exp\left(\frac{x}{1+x^2}\right)\frac{1-x^2}{(1+x^2)^2}.
$$
The <exponential function> and denominator are positive, so the <stationary points> occur exactly at $x=\pm1$. Their values are
$$
\boxed{f(-1)=e^{-1/2},\qquad f(1)=e^{1/2}.}
$$
The <derivative> is negative for $|x|>1$ and positive for $|x|<1$, so $x=-1$ is a <local minimum> and $x=1$ a <local maximum>. They are also global extrema, since $-1/2\leq x/(1+x^2)\leq1/2$.
Back to article page
