# What is the equation of the normal line of f(x)=e^(4x)/(4x) at x=-1?

Oct 14, 2016

$\textcolor{g r e e n}{y = \left(\frac{- 5 {e}^{- 4}}{4}\right) x - \frac{3 {e}^{- 4}}{2}}$

#### Explanation:

To find the equation of a straight line $\textcolor{g r e e n}{y = \textcolor{red}{a} x + b}$ we should have :
1) $\textcolor{red}{s l o p e = a}$
2)Point $\left(\textcolor{b r o w n}{{x}_{1} , {y}_{1}}\right)$ where the line passes through.

1)What is the $\textcolor{red}{s l o p e}$

Since this line is normal and tangent to $f \left(x\right)$ then the slope is by computing $\textcolor{red}{f ' \left(- 1\right)}$

Since $f \left(x\right)$ is a rational function of the form $\frac{u \left(x\right)}{v \left(x\right)}$ then its derivative is:
color(blue)(((u(x))/(v(x)))'=(u'(x)*v(x)-v'(x)*u(x))/(v^2(x))

$f \left(x\right) = {e}^{4 x} / \left(4 x\right)$
$f ' \left(x\right) = \textcolor{b l u e}{\frac{\left({e}^{4 x}\right) ' \cdot \left(4 x\right) - \left(4 x\right) ' \left({e}^{4 x}\right)}{4 x} ^ 2}$
$f ' \left(x\right) = \frac{4 {e}^{4 x} \cdot 4 x - 4 {e}^{4 x}}{16 {x}^{2}}$
$f ' \left(x\right) = \frac{16 x {e}^{4 x} - 4 {e}^{4 x}}{16 {x}^{2}}$

Simplify by $4$
$f ' \left(x\right) = \frac{4 x {e}^{4 x} - {e}^{4 x}}{4 {x}^{2}}$

$\textcolor{red}{f ' \left(- 1\right)} = \frac{4 \left(- 1\right) {e}^{4 \left(- 1\right)} - {e}^{4 \left(- 1\right)}}{4 {\left(- 1\right)}^{2}}$
$\textcolor{red}{f ' \left(- 1\right)} = \frac{- 4 {e}^{- 4} - {e}^{- 4}}{4}$
$\textcolor{red}{f ' \left(- 1\right)} = \frac{- 5 {e}^{- 4}}{4}$
$\textcolor{red}{s l o p e}$ $\textcolor{red}{a = \frac{- 5 {e}^{- 4}}{4}}$

2)What is the coordinates of the point$\left(\textcolor{b r o w n}{{x}_{1} , {y}_{1}}\right)$?

Since the line is normal at $x = - 1$ so it passes through this point of abscissa $- 1$ let us find its ordinate $f \left(- 1\right)$
$\textcolor{b r o w n}{{y}_{1}} = f \left(- 1\right) = \frac{{e}^{- 4}}{-} 4$
$\textcolor{b r o w n}{{y}_{1}} = - \frac{{e}^{- 4}}{4}$

Coordinates of the point are$\left(\textcolor{b r o w n}{- 1 , - \frac{{e}^{- 4}}{4}}\right)$

The equation of the normal line to $f \left(x\right)$is:
$y - \textcolor{b r o w n}{{y}_{1}} = \textcolor{red}{a} \cdot \left(x - \textcolor{b r o w n}{{x}_{1}}\right)$
$y - \left(- \frac{{e}^{- 4}}{4}\right) = \frac{- 5 {e}^{- 4}}{4} \left(x - \left(- 1\right)\right)$
$y + \frac{{e}^{- 4}}{4} = \frac{- 5 {e}^{- 4}}{4} \left(x + 1\right)$
$y + \frac{{e}^{- 4}}{4} = \left(\frac{- 5 {e}^{- 4}}{4}\right) x - \frac{5 {e}^{- 4}}{4}$
$y = \left(\frac{- 5 {e}^{- 4}}{4}\right) x - \frac{5 {e}^{- 4}}{4} - \frac{{e}^{- 4}}{4}$
$y = \left(\frac{- 5 {e}^{- 4}}{4}\right) x - \frac{6 {e}^{- 4}}{4}$
Simplify $\frac{6 {e}^{- 4}}{4}$ by $2$
Therefore ,Equation of the line is:
$\textcolor{g r e e n}{y = \left(\frac{- 5 {e}^{- 4}}{4}\right) x - \frac{3 {e}^{- 4}}{2}}$