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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 0, x = 2

Domain of

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

Find intervals:

Increasing

: 0 < x < 2

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 2

into

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

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

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

f (x) = 4 x^{2} - 16 x

2 - 2 x^{2}

f (x, y) = 10 x^{2} - 7 y^{2}

f (x) = - (\frac{4}{x ^{2}})

f (x) = x^{3} - 27 x + 63

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