What is the equation of the normal line of f(x)=1xe−x3+x2 at x=-1

Expert answers to "What is the equation of the normal line..."

Lilliana Villa

Answered question

2022-04-08

What is the equation of the normal line of

f (x) = \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^{'} = (- 3 x + 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^{'} = 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

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