What is the extreme f(x)=xe^{-(x^2)/(162)} ?

The extreme f(x)=xe^{-(x^2)/(162)} is Minimum(-9,-9/(e^{1/2)}),Maximum(9, 9/(e^{1/2)})

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extreme f(x)=xe^{-(x^2)/(162)}

extreme points

f (x) = x e^{- \frac{x ^{2}}{162}}

Minimum (- 9, - \frac{9}{e ^{\frac{1}{2}}}), Maximum (9, \frac{9}{e ^{\frac{1}{2}}})

Solution steps

Find the critical points:

x = - 9, x = 9

Domain of

x e^{- \frac{x ^{2}}{162}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 9,

Increasing

: - 9 < x < 9,

Decreasing

: 9 < x < \infty

Plug the extreme point

x = - 9

into

x e^{- \frac{x ^{2}}{162}} \Rightarrow y = - \frac{9}{e ^{\frac{1}{2}}}

Minimum (- 9, - \frac{9}{e ^{\frac{1}{2}}})

Plug the extreme point

x = 9

into

x e^{- \frac{x ^{2}}{162}} \Rightarrow y = \frac{9}{e ^{\frac{1}{2}}}

Maximum (9, \frac{9}{e ^{\frac{1}{2}}})

Minimum (- 9, - \frac{9}{e ^{\frac{1}{2}}}), Maximum (9, \frac{9}{e ^{\frac{1}{2}}})

f (x, y) = x^{2} + 2 \cdot y^{2} + 4 \cdot x - 4 \cdot y

f (x) = x^{4} - 8 x^{2} + 18

f (x) = x^{3} + 29 x + 128

x^{2} + \frac{4}{x}

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

What is the extreme f(x)=xe^{-(x^2)/(162)} ?
The extreme f(x)=xe^{-(x^2)/(162)} is Minimum(-9,-9/(e^{1/2)}),Maximum(9, 9/(e^{1/2)})