Question #64b09

Nov 12, 2016

$f \left(- 4\right) = 4 {e}^{\frac{1}{8}}$ is the maximum value of $f$ on $\left[- 4 , 16\right]$

and

$f \left(16\right) = - 16 {e}^{2}$ is the minimum value of $f$ on $\left[- 4 , 16\right]$

Explanation:

Given a continuous function $f \left(x\right)$ a closed interval $\left[a , b\right]$, any extrema will occur either at an endpoint or a critical point, that is, a point where $f ' \left(x\right) = 0$ or a point at which $f ' \left(x\right)$ does not exist.

In our case, we have $f \left(- x\right) = x {e}^{{\left(- x\right)}^{2} / 128}$, meaning $f \left(x\right) = - x {e}^{{x}^{2} / 128}$.

Differentiating, we can use the product rule and the chain rule to get

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} - x {e}^{{x}^{2} / 128}$

$= - x \left(\frac{d}{\mathrm{dx}} {e}^{{x}^{2} / 128}\right) + {e}^{{x}^{2} / 128} \left(\frac{d}{\mathrm{dx}} - x\right)$

$= - x \left[{e}^{{x}^{2} / 128} \left(\frac{d}{\mathrm{dx}} {x}^{2} / 128\right)\right] + {e}^{{x}^{2} / 128} \left(- 1\right)$

$= - x {e}^{{x}^{2} / 128} \left(\frac{x}{64}\right) - {e}^{{x}^{2} / 128}$

$= - {e}^{{x}^{2} / 128} \left({x}^{2} / 64 + 1\right)$

As this is defined for all real numbers, we only need look for points at which $f ' \left(x\right) = 0$. Doing so, we get

$- {e}^{{x}^{2} / 128} \left({x}^{2} / 64 + 1\right) = 0$

As ${e}^{{x}^{2} / 128} > 0$ for all real numbers, ${x}^{2} / 64 + 1 > {x}^{2} / 64 \ge 0$ for all real numbers, and the product of two nonzero values is nonzero, we know that there are no real solutions to the above. Thus, our function has no critical points.

As our function has no critical points on $\left[- 4 , 16\right]$ (or anywhere, for that matter), its extrema must occur at the endpoints. Evaluating, we get

$f \left(- 4\right) = 4 {e}^{{\left(- 4\right)}^{2} / 128} = 4 {e}^{\frac{1}{8}}$

$f \left(16\right) = - 16 {e}^{{16}^{2} / 128} = - 16 {e}^{2}$

So the above are the extrema of the function on $\left[- 4 , 16\right]$, giving us our answer:

$f \left(- 4\right) = 4 {e}^{\frac{1}{8}}$ is the maximum value of $f$ on $\left[- 4 , 16\right]$

and

$f \left(16\right) = - 16 {e}^{2}$ is the minimum value of $f$ on $\left[- 4 , 16\right]$