What is the equation of the normal line of $f (x) = \frac{x^{e}}{e^{x}}$ $f(x)= x^e/e^x$ at $x = e$ $x=e$ ?

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Answer 1 · 2017-05-11T09:23:02

$y = 1 .$ $y=1.$

$f (x) = \frac{x^{e}}{e^{x}} = x^{e} \cdot e^{- x} \Rightarrow f (e) = \frac{e^{e}}{e^{e}} = 1 .$ $f(x)=x^e/e^x=x^e*e^-x rArr f(e)=e^e/e^e=1.$

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

$(e, 1), say P (e, 1) .$ $(e,1)," say "P(e,1).$

$:. f'(x)=x^e*d/dx(e^-x)+e^-x*d/dx(x^e).$

$=x^e*(-e^-x)+e^-x*(ex^(e-1)).$

$:. f'(e)=e^e*(-e^-e)+e^-e*(ee^(e-1)).$

$=-e^e*1/e^e+e^(-e+1+e-1)$

$:. f'(e)=-1+e^0=-1+1=0.$

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 $P(e,1)$ to the curve is Horizontal.

Since the Normal at $P(e,1)" is "bot$ 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 $y=1.$

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