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

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

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

extreme points

f (x) = x^{2} \cdot 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} \cdot e^{- x} : 0

Minimum (0, 0)

Plug

x = 2

into

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

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

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

y = x^{2} - 4 x

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

f (x) = 3 x^{2} - 12 x + 9

f (x) = \frac{10}{3} x^{3} - x^{2} - 8 x + 48

f (x, y) = x y - 2 x - y

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