# How do you find the local extrema for (e^x)(x^2)?

Feb 20, 2017

#### Answer:

$\text{ Local Mninma is 0, Local Maxima is 4/e^2.}$

#### Explanation:

Suppose that, $f \left(x\right) = {x}^{2} {e}^{x} .$

For Local Maxima, f'(x)=0, &, f''(x)<0.

For Local Minima, f'(x)=0, &, f''(x)>0.

$f \left(x\right) = {x}^{2} {e}^{x} \therefore f ' \left(x\right) = {x}^{2} {e}^{x} + 2 x {e}^{x} = x \left(x + 2\right) {e}^{x} .$

$f ' \left(x\right) = \left({x}^{2} + 2 x\right) {e}^{x} , \therefore f ' ' \left(x\right) = \left({x}^{2} + 2 x\right) {e}^{x} + \left(2 x + 2\right) {e}^{x} .$

$\therefore f ' ' \left(x\right) = \left({x}^{2} + 4 x + 2\right) {e}^{x} .$

Now, $f ' \left(x\right) = 0 \Rightarrow x \left(x + 2\right) {e}^{x} = 0 \Rightarrow x = 0 , \mathmr{and} , x = - 2.$

$f ' ' \left(0\right) = 2 > 0 , \Rightarrow \text{ f has a local Minima at x=0, and, it is f(0)=0.}$

Similarly, $f ' ' \left(- 2\right) = - 2 {e}^{-} 2 < 0 \therefore f \left(- 2\right) = \frac{4}{e} ^ 2 \text{ is local Maxima.}$

Enjoy Maths.!