# Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 64

Determine the absolute maximum and absolute minimum values of $\displaystyle f(x) = x^2 e^{\frac{-x}{2}}$ on the interval $[-1,6]$

To determine the critical numbers, we set $f'(x) = 0$, so..

\begin{aligned} \text{if } f(x) =& x^2 e^{\frac{-x}{2}} \text{ then by using Product Rule..} \\ \\ f'(x) =& \left[ x^2 e^{\frac{-x}{2}} \left( \frac{-1}{2} \right) + 2x e^{\frac{-x}{2}} \right] \\ \\ f'(x) =& xe^{\frac{-x}{2}} \left( 2 - \frac{x}{2} \right) \end{aligned}

When $f'(x) = 0$, then..

$\displaystyle 0 = x e^{\frac{-x}{2}} \left( 2 - \frac{x}{2} \right)$

We have,

$xe^{\frac{-x}{2}} = 0$ and $\displaystyle 2 - \frac{x}{2} = 0$

The real solution and the critical number is..

\begin{aligned} 2 - \frac{x}{2} =& 0 \\ \\ \frac{x}{2} =& 2 \\ \\ x =& 4 \end{aligned}

So we have either a maximum or a minimum at $x = 4$, if we evaluate $f(x)$ with $x = 4$, the intervals $x = \pm 1$ and $x = 6$ and $x = 0$,

$\begin{array}{ccc} \text{when } x = 4, & & \text{when }x = -1, \\ f(4) = 4^2 (e^{\frac{-4}{2}}) & & f(-1) = (-1)^2 (e^{\frac{-4(-1)}{2}} ) \\ f(4) = 2.1654 & & f(-1) = 1.6487 \\ \text{when } x = 6,& & \text{when } x = 0,\\ f(6) = 6^2 (e^{\frac{-6}{2}}) & & f(0) = (0)^2 (e^{\frac{-0}{2}}) \\ f(6) = 1.7923 & & f(0) = 0 \end{array}$

Therefore, the absolute maximum is $f(4) = 2.1654$ and the absolute minimum is $f(0) = 0$.

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