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

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Essence Ingram

2022-04-07

What is the equation of the normal line of

f (x) = \frac{1}{x} - e^{- x^{3} + x^{2}}

at x=-1?

tabuevniru8op

Beginner2022-04-08Added 14 answers

Explanation:
To begin with , for x=-1, f(x) would be

- 1 - e^{1 + 1} \to - 1 - e^{2}

. So the normal has
to be found at point

(- 1, - 1 - e^{2})

Now slope of f(x) would be

f^{'} (x) = - \frac{1}{x^{2}} - e^{x^{2} - x^{3}} (2 x - 3 x^{2})

. The slope at x=-1
would be

f^{'} (- 1) = - 1 - e^{2} (- 2 - 3) = - 1 + 5 e^{2}

The slope of the normal line would thus be

\frac{1}{1 - 5 e^{2}}

The equation(point-slope form) of the normal line would thus be

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

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