What is the extreme f(x)=x^4e^{-2x} ?

The extreme f(x)=x^4e^{-2x} is Minimum(0,0),Maximum(2,(16)/(e^4))

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extreme f(x)=x^4e^{-2x}

extreme points

f (x) = x^{4} e^{- 2 x}

Minimum (0, 0), Maximum (2, \frac{16}{e ^{4}})

Solution steps

f^{'} (x) = 2 e^{- 2 x} x^{3} (2 - x)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

x^{4} e^{- 2 x} : 0

Minimum (0, 0)

Plug

x = 2

into

x^{4} e^{- 2 x} : \frac{16}{e ^{4}}

Maximum (2, \frac{16}{e ^{4}})

Minimum (0, 0), Maximum (2, \frac{16}{e ^{4}})

f (x) = e^{x^{2} - 8 x - 1}

f (x, y) = 36 - x^{2} - y^{2}

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

x 4 - x^{2}

f (x, y) = 4 x^{2} + 8 x + 4 y^{2} + 4

What is the extreme f(x)=x^4e^{-2x} ?
The extreme f(x)=x^4e^{-2x} is Minimum(0,0),Maximum(2,(16)/(e^4))