What is the extreme f(x)=x^2e^{-5x} ?

The extreme f(x)=x^2e^{-5x} is Minimum(0,0),Maximum(2/5 , 4/(25e^2))

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extreme f(x)=x^2e^{-5x}

extreme points

f (x) = x^{2} e^{- 5 x}

Minimum (0, 0), Maximum (\frac{2}{5}, \frac{4}{25 e ^{2}})

Solution steps

f^{'} (x) = e^{- 5 x} x (2 - 5 x)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{2}{5},

Decreasing

: \frac{2}{5} < x < \infty

Plug

x = 0

into

x^{2} e^{- 5 x} : 0

Minimum (0, 0)

Plug

x = \frac{2}{5}

into

x^{2} e^{- 5 x} : \frac{4}{25 e ^{2}}

Maximum (\frac{2}{5}, \frac{4}{25 e ^{2}})

Minimum (0, 0), Maximum (\frac{2}{5}, \frac{4}{25 e ^{2}})

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

7 y + 42 = - 14 x

x - 7 \cdot x - 2

y = - 1.4 x - 1

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

What is the extreme f(x)=x^2e^{-5x} ?
The extreme f(x)=x^2e^{-5x} is Minimum(0,0),Maximum(2/5 , 4/(25e^2))