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

The tangent of f(x)=xe^{-x/2},\at x= 1/2 is y= 3/(4e^{1/4)}x+1/(8e^{1/4)}

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tangent of f(x)=xe^{-x/2},\at x= 1/2

tangent of

f (x) = x e^{- \frac{x}{2}}, at x = \frac{1}{2}

y = \frac{3}{4 e ^{\frac{1}{4}}} x + \frac{1}{8 e ^{\frac{1}{4}}}

Solution steps

Find the tangent point:

(\frac{1}{2}, \frac{1}{2 e ^{\frac{1}{4}}})

Find the slope of

f (x) = x e^{- \frac{x}{2}} : \frac{df ( x )}{d x} = e^{- \frac{x}{2}} - \frac{e ^{- \frac{x}{2}} x}{2}

Compute the slope of

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

x = \frac{1}{2} : m = \frac{3}{4 e ^{\frac{1}{4}}}

Find the line with slope m=

\frac{3}{4 e ^{\frac{1}{4}}}

and passing through

(\frac{1}{2}, \frac{1}{2 e ^{\frac{1}{4}}}) : y = \frac{3}{4 e ^{\frac{1}{4}}} x + \frac{1}{8 e ^{\frac{1}{4}}}

y = \frac{3}{4 e ^{\frac{1}{4}}} x + \frac{1}{8 e ^{\frac{1}{4}}}

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t^{2} y^{^{''}} + 2 t y^{^{'}} - 30 y = 0

x \to 3 lim (\frac{( x + 3 )}{x - 3})

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What is the tangent of f(x)=xe^{-x/2},\at x= 1/2 ?
The tangent of f(x)=xe^{-x/2},\at x= 1/2 is y= 3/(4e^{1/4)}x+1/(8e^{1/4)}