What is the tangent of f(x)= x/(e^x),\at x=-1 ?

The tangent of f(x)= x/(e^x),\at x=-1 is y=2ex+e

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tangent of f(x)= x/(e^x),\at x=-1

tangent of

f (x) = \frac{x}{e ^{x}}, at x = - 1

y = 2 e x + e

Solution steps

Find the tangent point:

(- 1, - e)

Find the slope of

f (x) = \frac{x}{e ^{x}} : \frac{df ( x )}{d x} = \frac{1 - x}{e ^{x}}

Compute the slope of

f (x) = \frac{x}{e ^{x}}

x = - 1 : m = 2 e

Find the line with slope m=

2 e

and passing through

(- 1, - e) : y = 2 e x + e

y = 2 e x + e

y^{^{'}} = 8 x e^{(5 y)}, y (0) = 0

\frac{d}{d x} (arccot (2 x))

\int \frac{7 ( x ^{2} + 5 )}{x ( x ^{2} + 7 )} d x

3 x^{3} - x^{2} - 10 x, - x^{2} + 2 x, [- 2, 2]

\int_{- 2}^{0} 2 x^{2} + 5 x + 6 d x

What is the tangent of f(x)= x/(e^x),\at x=-1 ?
The tangent of f(x)= x/(e^x),\at x=-1 is y=2ex+e