Answer
Absolute maximum value is 1.6487 and the absolute minimum value is -0.7358.
Work Step by Step
$f(x) =xe^{\frac{x}{2}}$
Apply product rule of differentiation.
$f'(x) =e^{\frac{x}{2}}(1+\frac{x}{2})$
Now put, $f'(x) =0$
$e^{\frac{x}{2}}(1+\frac{x}{2})=0$
Since, $e^{\frac{x}{2}}\ne0$
Thus, $(1+\frac{x}{2})=0$
This implies $x=-2$
As we are given $f(x) =xe^{\frac{x}{2}}$ at intervals [-3,1]
$f(-3)=-3e^{\frac{-3}{2}}=-0.6694$
$f(-2)=-2e^{\frac{-2}{2}}=-0.7358$
$f(1)=1e^{\frac{1}{2}}=1.6487$
Therefore, the absolute maximum value at x =-3,-2,1 for
$f(x) =xe^{\frac{x}{2}}$ is 1.6487 and the absolute minimum value is -0.7358.
