Lilliana Villa

Answered question

2022-04-08

What is the equation of the normal line of $f\left(x\right)=\frac{1}{x}{e}^{-{x}^{3}+{x}^{2}}$ at x=-1

Answer & Explanation

gsmckibbenx7ga

Beginner2022-04-09Added 17 answers

Explanation:
You start by finding the first derivative .
${y}^{\prime }=\left(-3x+2\right){e}^{-{x}^{3}+{x}^{2}}$
by substituting with x=-1 in both the function and the first derivative you get the y co_ordinate of the point that lies on the normal line and the slope of the tangent to the curve ( m )
$y=\frac{1}{-1}{e}^{2}$
so the y coordinate of the point $=-{e}^{2}$
and the slope of the tangent ( m ) at x=-1 equals:
${y}^{\prime }=5{e}^{2}$
but We don't want the slope of the tangent, We want the slope of the normal.
The slope of the normal $=-\frac{1}{m}$
so it will be $-\frac{1}{5}{e}^{-2}$
and to get the equation of the straight line
$y-\left(-{e}^{2}\right)=-\frac{1}{5}{e}^{-2}\left(x-\left(-1\right)\right)$
and by simplification you get:
$5{e}^{2}y+5{e}^{4}+x+1=0$

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