What is the extreme f(x)= 1/2 x^2e^{-x} ?

The extreme f(x)= 1/2 x^2e^{-x} is Minimum(0,0),Maximum(2, 2/(e^2))

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extreme f(x)= 1/2 x^2e^{-x}

extreme points

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

Minimum (0, 0), Maximum (2, \frac{2}{e ^{2}})

Solution steps

Find the critical points:

x = 0, x = 2

Domain of

\frac{1}{2} x^{2} e^{- x} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug the extreme point

x = 0

into

\frac{1}{2} x^{2} e^{- x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 2

into

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

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

Minimum (0, 0), Maximum (2, \frac{2}{e ^{2}})

f (x, y) = 3 x y - x^{2} y - x y^{2}

f (x) = x^{4} - 12 x^{3} - 29

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

f (x) = \frac{x ^{2} + 1}{x ^{2} - 4}

f (x) = \frac{1}{3} x^{3} - 9 x + 2

What is the extreme f(x)= 1/2 x^2e^{-x} ?
The extreme f(x)= 1/2 x^2e^{-x} is Minimum(0,0),Maximum(2, 2/(e^2))