What is the extreme f(x)=xe^{-(x^2)/(32)} ?

The extreme f(x)=xe^{-(x^2)/(32)} is Minimum(-4,-4/(e^{1/2)}),Maximum(4, 4/(e^{1/2)})

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extreme f(x)=xe^{-(x^2)/(32)}

extreme points

f (x) = x e^{- \frac{x ^{2}}{32}}

Minimum (- 4, - \frac{4}{e ^{\frac{1}{2}}}), Maximum (4, \frac{4}{e ^{\frac{1}{2}}})

Solution steps

Find the critical points:

x = - 4, x = 4

Domain of

x e^{- \frac{x ^{2}}{32}} : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 4,

Decreasing

: 4 < x < \infty

Plug the extreme point

x = - 4

into

x e^{- \frac{x ^{2}}{32}} \Rightarrow y = - \frac{4}{e ^{\frac{1}{2}}}

Minimum (- 4, - \frac{4}{e ^{\frac{1}{2}}})

Plug the extreme point

x = 4

into

x e^{- \frac{x ^{2}}{32}} \Rightarrow y = \frac{4}{e ^{\frac{1}{2}}}

Maximum (4, \frac{4}{e ^{\frac{1}{2}}})

Minimum (- 4, - \frac{4}{e ^{\frac{1}{2}}}), Maximum (4, \frac{4}{e ^{\frac{1}{2}}})

x^{\frac{25}{3}} - 6 x^{\frac{19}{3}}

f (x) = \frac{x}{x - 2}, 3 \leq x \leq 4

f (x) = x^{2} + y^{2} - 6 x + 4 y - 10

2 x^{2} (1 - x^{2})

P (x, y) = - 4 x + 52 y - x^{2} + 2 x y - 7 y^{2} + 9

What is the extreme f(x)=xe^{-(x^2)/(32)} ?
The extreme f(x)=xe^{-(x^2)/(32)} is Minimum(-4,-4/(e^{1/2)}),Maximum(4, 4/(e^{1/2)})