Finding critical numbers

Using the quotient rule, $f'(x) = \dfrac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \dfrac{1 - x^2}{(x^2+1)^2}$. Since $x^2+1 \ne 0$ for all real $x$, $f'(x)$ is defined everywhere, so the only critical numbers come from setting the numerator equal to zero.

Solving $1 - x^2 = 0$ gives $x = 1$ and $x = -1$, both of which lie in $[-4, 4]$. Therefore the critical numbers are $x = -1$ and $x = 1$.

The candidates for extrema are the critical numbers $x = -1, 1$ and the endpoints $x = -4, 4$. Evaluating: $f(-4) = \dfrac{-4}{17} \approx -0.235$, $f(-1) = \dfrac{-1}{2} = -0.5$, $f(1) = \dfrac{1}{2} = 0.5$, and $f(4) = \dfrac{4}{17} \approx 0.235$.

Comparing these four values, the largest is $f(1) = 0.5$ and the smallest is $f(-1) = -0.5$. Therefore the absolute maximum value is $0.5$ at $x = 1$, and the absolute minimum value is $-0.5$ at $x = -1$.