What is the extreme f(x)=4x^2e^{-0.25x} ?

The extreme f(x)=4x^2e^{-0.25x} is Minimum(0,0),Maximum(8,(256)/(e^2))

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extreme f(x)=4x^2e^{-0.25x}

extreme

Minimum (0, 0), Maximum (8, \frac{256}{e ^{2}})

Solution steps

Find the critical points:

x = 0, x = 8

Domain of

4 x^{2} e^{- 0.25 x} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 8,

Decreasing

: 8 < x < \infty

Plug the extreme point

x = 0

into

4 x^{2} e^{- 0.25 x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 8

into

4 x^{2} e^{- 0.25 x} \Rightarrow y = \frac{256}{e ^{2}}

Maximum (8, \frac{256}{e ^{2}})

Minimum (0, 0), Maximum (8, \frac{256}{e ^{2}})

What is the extreme f(x)=4x^2e^{-0.25x} ?
The extreme f(x)=4x^2e^{-0.25x} is Minimum(0,0),Maximum(8,(256)/(e^2))