What is the extreme y=xe^{-(x^2)/(18)} ?

The extreme y=xe^{-(x^2)/(18)} is Minimum(-3,-3/(e^{1/2)}),Maximum(3, 3/(e^{1/2)})

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extreme y=xe^{-(x^2)/(18)}

extreme points

y = x e^{- \frac{x ^{2}}{18}}

Minimum (- 3, - \frac{3}{e ^{\frac{1}{2}}}), Maximum (3, \frac{3}{e ^{\frac{1}{2}}})

Solution steps

Find the critical points:

x = - 3, x = 3

Domain of

x e^{- \frac{x ^{2}}{18}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 3,

Decreasing

: 3 < x < \infty

Plug the extreme point

x = - 3

into

x e^{- \frac{x ^{2}}{18}} \Rightarrow y = - \frac{3}{e ^{\frac{1}{2}}}

Minimum (- 3, - \frac{3}{e ^{\frac{1}{2}}})

Plug the extreme point

x = 3

into

x e^{- \frac{x ^{2}}{18}} \Rightarrow y = \frac{3}{e ^{\frac{1}{2}}}

Maximum (3, \frac{3}{e ^{\frac{1}{2}}})

Minimum (- 3, - \frac{3}{e ^{\frac{1}{2}}}), Maximum (3, \frac{3}{e ^{\frac{1}{2}}})

f (x, y) = 2 x^{3} + 2 x y^{2} - 4 x + 5

f (x) = \frac{40}{r} + 4 π r^{2}

f (x) = x \cdot x + 2

f (x, y) = 13 - 4 x + 8 y

What is the extreme y=xe^{-(x^2)/(18)} ?
The extreme y=xe^{-(x^2)/(18)} is Minimum(-3,-3/(e^{1/2)}),Maximum(3, 3/(e^{1/2)})