Answer
The absolute maximum is $f(2) = \frac{2}{5}$
The absolute minimum is $f(-\frac{1}{3}) = -\frac{9}{2}$
There is no local maximum.
$f(-\frac{1}{3}) = -\frac{9}{2}$ is a local minimum.
Work Step by Step
$f(x) = \frac{3x-4}{x^2+1}$
Note that this function is continuous on the interval $[-2,2]$ and differentiable on the interval $(-2,2)$.
We can find the points where $f'(x) = 0$:
$f'(x) = \frac{3(x^2+1)-2x(3x-4)}{(x^2+1)^2}=0$
$3x^2+3-6x^2+8x=0$
$-3x^2+8x+3=0$
$3x^2-8x-3=0$
$(3x+1)(x-3)=0$
$x = -\frac{1}{3}, 3$
We can verify the value of the function when $f'(x) = 0$ and at the endpoints of the interval $[-2,2]$:
$f(-2) = \frac{3(-2)-4}{(-2)^2+1} = -2$
$f(-\frac{1}{3}) = \frac{3(-\frac{1}{3})-4}{(-\frac{1}{3})^2+1} = -\frac{9}{2}$
$f(2) = \frac{3(2)-4}{(2)^2+1} = \frac{2}{5}$
The absolute maximum is $f(2) = \frac{2}{5}$
The absolute minimum is $f(-\frac{1}{3}) = -\frac{9}{2}$
There is no local maximum.
$f(-\frac{1}{3}) = -\frac{9}{2}$ is a local minimum.