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

##### 1 Answer

$y = - \frac{x}{20 \cdot {e}^{2}} + \frac{1}{10 \cdot {e}^{2}} + 8 \cdot {e}^{2}$

#### Explanation:

From the given equation $f \left(x\right) = {x}^{3} \cdot {e}^{x}$ at $x = 2$

Let us solve for the point first. We should find $f \left(2\right)$

$f \left(2\right) = {2}^{3} \cdot {e}^{2} = 8 \cdot {e}^{2}$
We now have the point at $\left(2 , 8 {e}^{2}\right)$

Now, we need to compute the slope m and then obtain its negative reciprocal because, we are after the normal line.

First derivative of $f \left(x\right)$ is $f ' \left(x\right)$

$\frac{d}{\mathrm{dx}} f \left(x\right) = f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} \cdot {e}^{x}\right) = {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{3}\right) + {x}^{3} \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)$

$f ' \left(x\right) = 3 {x}^{2} \cdot {e}^{x} + {x}^{3} \cdot {e}^{x}$

for slope $m = f ' \left(2\right)$ use $x = 2$

$f ' \left(x\right) = 3 {x}^{2} \cdot {e}^{x} + {x}^{3} \cdot {e}^{x}$
$f ' \left(2\right) = 3 {\left(2\right)}^{2} \cdot {e}^{2} + {\left(2\right)}^{3} \cdot {e}^{2}$
$f ' \left(2\right) = 12 \cdot {e}^{2} + 8 \cdot {e}^{2}$
$f ' \left(2\right) = 20 \cdot {e}^{2}$
and $m = f ' \left(2\right) = 20 \cdot {e}^{2}$

For the normal line, we need $m ' = - \frac{1}{m}$

$m ' = - \frac{1}{m} = - \frac{1}{20 \cdot {e}^{2}}$

Let us write the Normal line equation using $m ' = - \frac{1}{20 \cdot {e}^{2}}$ and the point $\left(2 , 8 {e}^{2}\right)$

By Point-Slope form

$y - {y}_{1} = m ' \left(x - {x}_{1}\right)$

$y - 8 {e}^{2} = - \frac{1}{20 \cdot {e}^{2}} \left(x - 2\right)$

$y = - \frac{1}{20 \cdot {e}^{2}} \left(x - 2\right) + 8 {e}^{2}$

$\textcolor{red}{y = - \frac{x}{20 \cdot {e}^{2}} + \frac{1}{10 \cdot {e}^{2}} + 8 \cdot {e}^{2} \text{ }}$the required Normal Line

God bless....I hope the explanation is useful.