# What are the critical points of g(x)=e^(x^2)(x-10)?

Aug 21, 2017

#### Answer:

$x = \frac{2 \pm \sqrt{23}}{2}$

#### Explanation:

Domain of $g$ is $\mathbb{R}$

$g ' \left(x\right) = 2 x {e}^{{x}^{2}} \left(x - 10\right) + {e}^{{x}^{2}}$

$= {e}^{{x}^{2}} \left(2 {x}^{2} - 10 x + 1\right)$

$g ' \left(x\right)$ is never undefined and $g ' \left(x\right) = 0$ at

$2 {x}^{2} - 10 x + 1 = 0$

$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \left(2\right) \left(1\right)}}{2 \left(2\right)}$

$= \frac{10 \pm \sqrt{92}}{4} = \frac{10 \pm 2 \sqrt{23}}{4} = \frac{2 \pm \sqrt{23}}{2}$