# What is the equation of the line that is normal to f(x)= x^3e^sqrtx  at  x= 2 ?

Dec 2, 2016

$y - 8 {e}^{\sqrt{2}} = \frac{- 1}{\left[2 {e}^{\sqrt{2}} \left({2}^{\frac{1}{2}} + 6\right)\right]} \left(x - 2\right)$
$y = \frac{- x}{2 {e}^{\sqrt{2}} \left(\sqrt{2} + 6\right)} + \frac{1}{{e}^{\sqrt{2}} \left(\sqrt{2} + 6\right)} + 8 {e}^{\sqrt{2}}$

#### Explanation:

First, find the equation of the tangent line to $f \left(x\right)$ at x=2 in point slope form, then write the normal line equation the same, except with an opposite reciprocal slope.

To find slope of tangent line:
Find $f ' \left(x\right)$:
$f \left(x\right) = {x}^{3} {e}^{\sqrt{x}}$
$f ' \left(x\right) = \left({x}^{3}\right) \left(\frac{1}{2} {x}^{- \frac{1}{2}} {e}^{\sqrt{x}}\right) + \left(3 {x}^{2}\right) \left({e}^{\sqrt{x}}\right)$
$f ' \left(x\right) = \frac{{x}^{3} {e}^{\sqrt{x}}}{2 \sqrt{x}} + 3 {x}^{2} {e}^{\sqrt{x}}$
$f ' \left(x\right) = \left({x}^{2} {e}^{\sqrt{x}}\right) \left(\frac{1}{2} {x}^{\frac{1}{2}} + 3\right)$
$f ' \left(x\right) = \left(\frac{1}{2} {x}^{2} {e}^{\sqrt{x}}\right) \left({x}^{\frac{1}{2}} + 6\right)$

$f ' \left(2\right) = \left(\frac{1}{2} {\left(2\right)}^{2} {e}^{\sqrt{2}}\right) \left({\left(2\right)}^{\frac{1}{2}} + 6\right)$
$f ' \left(2\right) = \left[2 {e}^{\sqrt{2}} \left({2}^{\frac{1}{2}} + 6\right)\right]$

Therefore, slope of normal line:
$\frac{- 1}{f ' \left(2\right)} = \frac{- 1}{\left[2 {e}^{\sqrt{2}} \left({2}^{\frac{1}{2}} + 6\right)\right]}$

Point:
$f \left(2\right) = {2}^{3} {e}^{\sqrt{2}}$
$= 8 {e}^{\sqrt{2}}$
$\left(2 , 8 {e}^{\sqrt{2}}\right)$

Equation of normal line:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 8 {e}^{\sqrt{2}} = \frac{- 1}{\left[2 {e}^{\sqrt{2}} \left({2}^{\frac{1}{2}} + 6\right)\right]} \left(x - 2\right)$
$y = \frac{- 1}{2 {e}^{\sqrt{2}} \left({2}^{\frac{1}{2}} + 6\right)} \left(x - 2\right) + 8 {e}^{\sqrt{2}}$
$y = \frac{- x}{2 {e}^{\sqrt{2}} \left(\sqrt{2} + 6\right)} + \frac{1}{{e}^{\sqrt{2}} \left(\sqrt{2} + 6\right)} + 8 {e}^{\sqrt{2}}$