What is the extreme f(x)=-xe^{-x/5} ?

The extreme f(x)=-xe^{-x/5} is Minimum(5,-5/e)

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extreme f(x)=-xe^{-x/5}

extreme points

f (x) = - x e^{- \frac{x}{5}}

Minimum (5, - \frac{5}{e})

Solution steps

Find the critical points:

x = 5

Domain of

- x e^{- \frac{x}{5}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 5,

Increasing

: 5 < x < \infty

Plug the extreme point

x = 5

into

- x e^{- \frac{x}{5}} \Rightarrow y = - \frac{5}{e}

Minimum (5, - \frac{5}{e})

V, V = \frac{1}{3} π r 2 h

f (x) = e^{x} \cdot x^{2}

3 x^{2} + y^{2} - 2 x y + 42 - 8 y

f (x) = - \frac{56}{81 ( x + 1 ) ^{\frac{10}{3}}}

f (x) = \frac{- 1}{x - 2}

What is the extreme f(x)=-xe^{-x/5} ?
The extreme f(x)=-xe^{-x/5} is Minimum(5,-5/e)