What is the extreme f(x)=4xe^{-x},0<= x<= 2 ?

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

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extreme f(x)=4xe^{-x},0<= x<= 2

extreme

Minimum (0, 0), Maximum (1, \frac{4}{e}), Minimum (2, \frac{8}{e ^{2}})

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

4 x e^{- x}, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 1

into

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

Maximum (1, \frac{4}{e})

Plug the extreme point

x = 2

into

4 x e^{- x} \Rightarrow y = \frac{8}{e ^{2}}

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

Minimum (0, 0), Maximum (1, \frac{4}{e}), Minimum (2, \frac{8}{e ^{2}})

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