What is the extreme f(x)=x^2e^{-3x} ?

The extreme f(x)=x^2e^{-3x} is Minimum(0,0),Maximum(2/3 , 4/(9e^2))

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

extreme points

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

Minimum (0, 0), Maximum (\frac{2}{3}, \frac{4}{9 e ^{2}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{2}{3},

Decreasing

: \frac{2}{3} < x < \infty

Plug

x = 0

into

x^{2} e^{- 3 x} : 0

Minimum (0, 0)

Plug

x = \frac{2}{3}

into

x^{2} e^{- 3 x} : \frac{4}{9 e ^{2}}

Maximum (\frac{2}{3}, \frac{4}{9 e ^{2}})

Minimum (0, 0), Maximum (\frac{2}{3}, \frac{4}{9 e ^{2}})

\frac{x ^{2}}{x - 8}

f (x) = \frac{- 3}{x ^{2} - 4}

f (x, y) = 2 x + 3 y

f (x) = x 9 - x

f (x) = \frac{x}{x ^{2} - x + 16}

What is the extreme f(x)=x^2e^{-3x} ?
The extreme f(x)=x^2e^{-3x} is Minimum(0,0),Maximum(2/3 , 4/(9e^2))