What is the extreme f(x)=x(e^{((-x^2))/8}) ?

The extreme f(x)=x(e^{((-x^2))/8}) is Minimum(-2,-2/(e^{1/2)}),Maximum(2, 2/(e^{1/2)})

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extreme f(x)=x(e^{((-x^2))/8})

extreme points

f (x) = x (e^{\frac{( - x ^{2} )}{8}})

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

Solution steps

Find the critical points:

x = - 2, x = 2

Domain of

x e^{\frac{- x ^{2}}{8}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug the extreme point

x = - 2

into

x e^{\frac{- x ^{2}}{8}} \Rightarrow y = - \frac{2}{e ^{\frac{1}{2}}}

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

Plug the extreme point

x = 2

into

x e^{\frac{- x ^{2}}{8}} \Rightarrow y = \frac{2}{e ^{\frac{1}{2}}}

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

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

f (x) = 3 x - 4

f (x) = x^{8} (x - 3)^{7}

f (x) = 3 (x + 2)^{2} - 12 = 0

f (x) = 2 x^{3} - 24 x - 4

What is the extreme f(x)=x(e^{((-x^2))/8}) ?
The extreme f(x)=x(e^{((-x^2))/8}) is Minimum(-2,-2/(e^{1/2)}),Maximum(2, 2/(e^{1/2)})