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

The extreme f(x)=x^2e^{-x} is Minimum(0,0),Maximum(2, 4/(e^2))

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

extreme points

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

Minimum (0, 0), Maximum (2, \frac{4}{e ^{2}})

Solution steps

f^{'} (x) = e^{- x} x (2 - x)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

x^{2} e^{- x} : 0

Minimum (0, 0)

Plug

x = 2

into

x^{2} e^{- x} : \frac{4}{e ^{2}}

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

Minimum (0, 0), Maximum (2, \frac{4}{e ^{2}})

y = 13 x + 2

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

y = 4 x - 8

(1, 3), (4, - 3)

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

What is the extreme f(x)=x^2e^{-x} ?
The extreme f(x)=x^2e^{-x} is Minimum(0,0),Maximum(2, 4/(e^2))