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

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

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

extreme points

f (x) = x e^{- 4 x}, 0 \leq x \leq 2

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

Solution steps

Find the critical points:

x = 0, x = \frac{1}{4}, x = 2

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < 2

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{4}

into

x e^{- 4 x} \Rightarrow y = \frac{1}{4 e}

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

Plug the extreme point

x = 2

into

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

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

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

f (x) = - 1 - x + x^{2}

f (x) = - x^{2} + 4, - 2 \leq x \leq 3

x^{\frac{1}{5}} (x + 6)

f (x) = x^{2} e^{4 x}

f (x) = (x + 1)^{5} - 5 x - 2

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