What is the extreme f(x)=xe^{-x}[0.2] ?

The extreme f(x)=xe^{-x}[0.2] is Maximum(1,(0.2)/e)

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extreme f(x)=xe^{-x}[0.2]

extreme points

f (x) = x e^{- x} [0.2]

Maximum (1, \frac{0.2}{e})

Solution steps

Find the critical points:

x = 1

Domain of

x e^{- x} [0.2] : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \infty

Plug the extreme point

x = 1

into

x e^{- x} [0.2] \Rightarrow y = \frac{0.2}{e}

Maximum (1, \frac{0.2}{e})

f (x) = - 2 x^{2} - 12 x - 23

x^{2} - x y + y^{2} + 3 x - 2 y - 5

h (x) = \frac{χ ^{2} + 4 x - 32}{x ^{2} - 8 x + 16}

f (x) = 2 x^{3} 6 x^{2} - 18 x + 1

f (x) = 2 x^{3} - y^{2} - 6 x + 9 y

What is the extreme f(x)=xe^{-x}[0.2] ?
The extreme f(x)=xe^{-x}[0.2] is Maximum(1,(0.2)/e)