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

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

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

extreme points

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

Minimum (0, 0), Maximum (1, \frac{1}{e ^{2}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 1

into

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

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

Minimum (0, 0), Maximum (1, \frac{1}{e ^{2}})

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

f (x, y) = 2 x y - x^{2} - 2 y^{2} + 3 x + 4

f (x) = - x^{3} + 3 x^{2} + 45 x - 42

f (x) = (x^{2} + 12) (1 - x^{2})

\frac{x ^{2}}{x - 2}

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