Lilliana Villa

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

gsmckibbenx7ga

Beginner2022-04-09Added 17 answers

Explanation:

You start by finding the first derivative .

$y}^{\prime}=(-3x+2){e}^{-{x}^{3}+{x}^{2}$

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

$y-(-{e}^{2})=-\frac{1}{5}{e}^{-2}(x-(-1))$

and by simplification you get:

$5{e}^{2}y+5{e}^{4}+x+1=0$

You start by finding the first derivative .

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 )

so the y coordinate of the point

and the slope of the tangent ( m ) at x=-1 equals:

but We don't want the slope of the tangent, We want the slope of the normal.

The slope of the normal

so it will be

and to get the equation of the straight line

and by simplification you get:

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