What is the extreme f(x)=xe^{-(x^2)/(72)} ?

The extreme f(x)=xe^{-(x^2)/(72)} is Minimum(-6,-6/(e^{1/2)}),Maximum(6, 6/(e^{1/2)})

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extreme f(x)=xe^{-(x^2)/(72)}

extreme points

f (x) = x e^{- \frac{x ^{2}}{72}}

Minimum (- 6, - \frac{6}{e ^{\frac{1}{2}}}), Maximum (6, \frac{6}{e ^{\frac{1}{2}}})

Solution steps

Find the critical points:

x = - 6, x = 6

Domain of

x e^{- \frac{x ^{2}}{72}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 6,

Decreasing

: 6 < x < \infty

Plug the extreme point

x = - 6

into

x e^{- \frac{x ^{2}}{72}} \Rightarrow y = - \frac{6}{e ^{\frac{1}{2}}}

Minimum (- 6, - \frac{6}{e ^{\frac{1}{2}}})

Plug the extreme point

x = 6

into

x e^{- \frac{x ^{2}}{72}} \Rightarrow y = \frac{6}{e ^{\frac{1}{2}}}

Maximum (6, \frac{6}{e ^{\frac{1}{2}}})

Minimum (- 6, - \frac{6}{e ^{\frac{1}{2}}}), Maximum (6, \frac{6}{e ^{\frac{1}{2}}})

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y = x^{5} - 15 x^{3} + 100

f (x) = (ln (x^{2} + 1))

f (x) = 3 x + 5 x^{- 1}

What is the extreme f(x)=xe^{-(x^2)/(72)} ?
The extreme f(x)=xe^{-(x^2)/(72)} is Minimum(-6,-6/(e^{1/2)}),Maximum(6, 6/(e^{1/2)})