What is the tangent of f(x)=xe^{-x},\at x=2 ?

The tangent of f(x)=xe^{-x},\at x=2 is y=-1/(e^2)x+4/(e^2)

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tangent of f(x)=xe^{-x},\at x=2

tangent of

f (x) = x e^{- x}, at x = 2

y = - \frac{1}{e ^{2}} x + \frac{4}{e ^{2}}

Solution steps

Find the tangent point:

(2, \frac{2}{e ^{2}})

Find the slope of

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

Compute the slope of

f (x) = x e^{- x}

x = 2 : m = - \frac{1}{e ^{2}}

Find the line with slope m=

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

and passing through

(2, \frac{2}{e ^{2}}) : y = - \frac{1}{e ^{2}} x + \frac{4}{e ^{2}}

y = - \frac{1}{e ^{2}} x + \frac{4}{e ^{2}}

\frac{( s + 1 )}{s + 2}

y = 3 sec (x)

\frac{d}{d x} (uv)

\int \frac{x ^{2}}{1 - x ^{3}} d x

y = 8 - x^{2}, y = x^{2}

What is the tangent of f(x)=xe^{-x},\at x=2 ?
The tangent of f(x)=xe^{-x},\at x=2 is y=-1/(e^2)x+4/(e^2)