Finding critical numbers of f

Apply the quotient rule: $f'(x) = \dfrac{(1)(x^2+1) - x(2x)}{(x^2+1)^2} = \dfrac{1 - x^2}{(x^2+1)^2}$. The denominator $(x^2+1)^2$ is always positive and never zero, so $f'(x)$ is defined for all $x$.

Setting the numerator equal to zero: $1 - x^2 = 0$, so $x = \pm 1$. These are the only critical numbers since $f'(x)$ is never undefined.

The critical numbers $x = -1$ and $x = 1$ divide the real line into three intervals: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. Since $(x^2+1)^2 > 0$ always, the sign of $f'(x)$ matches the sign of $1 - x^2$.

For $x$ in $(-\infty,-1)$, $1-x^2 < 0$, so $f' < 0$ (decreasing). For $x$ in $(-1,1)$, $1-x^2 > 0$, so $f' > 0$ (increasing). For $x$ in $(1,\infty)$, $1-x^2 < 0$, so $f' < 0$ (decreasing).

At $x=-1$, $f'$ changes from negative (on $(-\infty,-1)$) to positive (on $(-1,1)$), so by the First Derivative Test $x=-1$ gives a local minimum. Compute $f(-1) = \dfrac{-1}{(-1)^2+1} = -\dfrac{1}{2}$.

At $x=1$, $f'$ changes from positive (on $(-1,1)$) to negative (on $(1,\infty)$), so $x=1$ gives a local maximum. Compute $f(1) = \dfrac{1}{1^2+1} = \dfrac{1}{2}$.