## Calculus 8th Edition

$f(x) =xe^{\frac{x}{2}}$ 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.