Answer
$$
f(x)=\frac{3 x-4}{x^{2}+1}, \quad [-2,2],
$$
The local minimum valuei s
$$
f\left(-\frac{1}{3}\right)=\frac{-5}{10 / 9}=-\frac{9}{2}
$$
is a local minimum value.
Checking the endpoints, we find
$$
f(-2)=-2 \text { and } f(2)=\frac{2}{5} .
$$
Thus,
$$
f\left(-\frac{1}{3}\right)=-\frac{9}{2}
$$
is the absolute minimum value and
$$
f(2)=\frac{2}{5}
$$
is the absolute maximum value
Work Step by Step
$$
f(x)=\frac{3 x-4}{x^{2}+1}, \quad [-2,2],
$$
Since $ f$ is continuous on [-2,2],we can use the Closed Interval Method
$$
f(x)=\frac{3 x-4}{x^{2}+1}
$$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{\left(x^{2}+1\right)(3)-(3 x-4)(2 x)}{\left(x^{2}+1\right)^{2}} \\
& =\frac{-\left(3 x^{2}-8 x-3\right)}{\left(x^{2}+1\right)^{2}} \\
&=\frac{-(3 x+1)(x-3)}{\left(x^{2}+1\right)^{2}}
\end{aligned}
$$
Since $f^{\prime}(x)$ exists for all $x$ the only critical numbers of $f$ occur when $f^{\prime}(x)=0$ that is,
$$
x=-\frac{1}{3} \text { or } x=3
$$
but $3$ is not in the interval.
For $-\frac{1}{3}0 $$
For $ -2
