What is the extreme f(x)=x^2e^{-8x} ?

The extreme f(x)=x^2e^{-8x} is Minimum(0,0),Maximum(1/4 , 1/(16e^2))

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

extreme points

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

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{1}{16 e ^{2}})

Solution steps

f^{'} (x) = 2 e^{- 8 x} x (1 - 4 x)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < \infty

Plug

x = 0

into

x^{2} e^{- 8 x} : 0

Minimum (0, 0)

Plug

x = \frac{1}{4}

into

x^{2} e^{- 8 x} : \frac{1}{16 e ^{2}}

Maximum (\frac{1}{4}, \frac{1}{16 e ^{2}})

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{1}{16 e ^{2}})

What is the extreme f(x)=x^2e^{-8x} ?
The extreme f(x)=x^2e^{-8x} is Minimum(0,0),Maximum(1/4 , 1/(16e^2))