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

The extreme f(x)=2x^2e^{-x/5} is Minimum(0,0),Maximum(10,(200)/(e^2))

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

extreme points

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

Minimum (0, 0), Maximum (10, \frac{200}{e ^{2}})

Solution steps

Find the critical points:

x = 0, x = 10

Domain of

2 x^{2} e^{- \frac{x}{5}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 10,

Decreasing

: 10 < x < \infty

Plug the extreme point

x = 0

into

2 x^{2} e^{- \frac{x}{5}} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 10

into

2 x^{2} e^{- \frac{x}{5}} \Rightarrow y = \frac{200}{e ^{2}}

Maximum (10, \frac{200}{e ^{2}})

Minimum (0, 0), Maximum (10, \frac{200}{e ^{2}})

f (x, y) = 3 x y + x^{3} - y^{3} + 1

f (x) = 9 - 2 x + 4 y - x^{2} - 2 y^{3}

y = x (6 - 2 x)^{2}

\frac{x + 2}{x ^{3}}

f (x) = - 0.03 x^{2} + 300 x - 100000

What is the extreme f(x)=2x^2e^{-x/5} ?
The extreme f(x)=2x^2e^{-x/5} is Minimum(0,0),Maximum(10,(200)/(e^2))