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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Domain of

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

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

Minimum (0, 0)

Plug the extreme point

x = 1

into

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

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

Plug the extreme point

x = 2

into

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

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

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

y = ∣ x + 4 ∣ + \frac{4}{∣ x ∣ - 3}

6 e^{x - 4}

f (x) = \frac{9 x}{x ^{2} + 6}

f (x) = x^{2} - 6 x + 9 - 4 y^{2}

f (x) = x^{\frac{8}{7}} + 8 x^{\frac{1}{7}}

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