# What is the equation of the line normal to  f(x)=e^(sqrtx-x) at  x=4?

Sep 14, 2016

I get $y = \frac{4 {e}^{2}}{3} x - \frac{16 {e}^{2}}{3} + \frac{1}{e} ^ 2$

#### Explanation:

$f \left(x\right) = {e}^{\sqrt{x} - x}$

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

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

$f ' \left(4\right) = {e}^{\sqrt{4} - 4} \left(\frac{1}{2 \sqrt{4}} - 1\right) = {e}^{2 - 4} \left(\frac{1}{4} - 1\right) = {e}^{- 2} \left(- \frac{3}{4}\right) = - \frac{3}{4 {e}^{2}}$

The normal line it perpendicular to the tangent, so the slope of the normal line is $- \frac{1}{f ' \left(4\right)} = \frac{4 {e}^{2}}{3}$.

The line through $\left(4 , \frac{1}{e} ^ 2\right)$ with slope $m = \frac{4 {e}^{2}}{3}$ is

$y - \frac{1}{e} ^ 2 = \frac{4 {e}^{2}}{3} \left(x - 4\right)$ or

$y = \frac{4 {e}^{2}}{3} x - \frac{16 {e}^{2}}{3} + \frac{1}{e} ^ 2$