Answer
Absolute maximum: $(3,2/3)$
Absolute minimum: $(-3,-2/3)$
Work Step by Step
$ f(x)=\displaystyle \frac{4x}{x^{2}+9} \quad$ ... defined on $[-4,4].$
$ f^{\prime}(x)=\displaystyle \frac{4(x^{2}+9)-4x(2x)}{(x^{2}+9)^{2}}\quad$ ... quotient rule
$=\displaystyle \frac{-4x^{2}+36}{(x^{2}+9)^{2}}\quad$ ... defined on $(-4,4).$
$f^{\prime}(x)=0 \quad$ for
$-4x^{2}+36=0$
$x^{2}-9=0$
$(x+3)(x-3)=0$
$x=\pm 3\in(-4,4)\quad $(critical numbers)
$\left[\begin{array}{lccccccc}
x, \text{ interval} & -4 & (-4,-3) & -3 & (-3,3) & 3 & (3,4) & 4\\
t=\text{ test number} & & -3.5 & & 0 & & 3.5 & \\
\text{ sign of }f^{\prime}(t) & & - & & + & & - & \\
f(x) & -16/25 & \searrow & -2/3 & \nearrow & 2/3 & \searrow & 16/25 \\
& & & min & & max & &
\end{array}\right]$
Absolute maximum: $(3,2/3)$
Absolute minimum: $(-3,-2/3)$
