# What are the absolute extrema of  f(x)= xe^(x^2)/128in [-5,16]?

Nov 9, 2016

$f \left(16\right) = {e}^{256} / 8 \mathmr{and} f \left(0\right) = 0$

#### Explanation:

$f ' = \left(\frac{1}{128}\right) \left(x {e}^{{x}^{2}}\right) ' = \left(\frac{1}{128}\right) \left(\left(x\right) ' {e}^{{x}^{2}} + \left({e}^{{x}^{2}}\right) ' x\right) = \left(\frac{1}{128}\right) {e}^{{x}^{2}} \left(1 + 2 {x}^{2}\right)$

Here, ${e}^{{x}^{2}} \ge 1 \mathmr{and} 1 + 2 {x}^{2} \ge 1$. So, f'>=1>0#

And so, f is an increasing function in $x \in \left(- \infty , \infty\right)$

As $| f \left(- 5\right) | = \left(\frac{5}{128}\right) {e}^{25} < f \left(16\right) = {e}^{256} / 8$,

the absolute maximum =$f \left(16\right) = {e}^{256} / 8$ and,

as $| f | \ge 0 \mathmr{and} f \left(0\right) = 0$,

the absolute minimum=$f \left(0\right) = 0$.