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

May 11, 2017

$y = 1.$

#### Explanation:

$f \left(x\right) = {x}^{e} / {e}^{x} = {x}^{e} \cdot {e}^{-} x \Rightarrow f \left(e\right) = {e}^{e} / {e}^{e} = 1.$

Thus, we require the eqn. of Normal to the Curve at the point

$\left(e , 1\right) , \text{ say } P \left(e , 1\right) .$

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

$= {x}^{e} \cdot \left(- {e}^{-} x\right) + {e}^{-} x \cdot \left(e {x}^{e - 1}\right) .$

$\therefore f ' \left(e\right) = {e}^{e} \cdot \left(- {e}^{-} e\right) + {e}^{-} e \cdot \left(e {e}^{e - 1}\right) .$

$= - {e}^{e} \cdot \frac{1}{e} ^ e + {e}^{- e + 1 + e - 1}$

$\therefore f ' \left(e\right) = - 1 + {e}^{0} = - 1 + 1 = 0.$

This means that, the Slope of Tangent at the point $x = e$ to

the given curve is $0.$

In other words, the tgt. line at $P \left(e , 1\right)$ to the curve is Horizontal.

Since the Normal at $P \left(e , 1\right) \text{ is } \bot$ to the tgt. at that point, we

find that the reqd. normal is a Vertical line through P.

Therefore, the eqn. of the normal is $y = 1.$