Answer
Absolute maximum: $(0,0)$
Absolute minimum: $(-5/2,-25/4)$
Work Step by Step
$ f(x)=x^{2}+5x \quad$ ... defined everywhere
$ f^{\prime}(x)=2x+5\quad$ ... defined everywhere.
$f^{\prime}(x)=0 \quad$ for $x=-5/2\in[-4,0]\quad $(critical number)
$\left[\begin{array}{lccccc}
x, \text{ interval} & -4 & (-4,-5/2) & -5/2 & (-5/2,0) & 0\\
t=\text{ test number} & & -3 & & -1 & \\
f^{\prime}(t) & & -1 & & +3 & \\
f(x) & -4 & \searrow & -25/4 & \nearrow & 0\\
& & & min & & max
\end{array}\right]$
Absolute maximum: $(0,0)$
Absolute minimum: $(-5/2,-25/4)$
