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)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

f (x) = x \cdot 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 \cdot 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 \cdot 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 \cdot 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) = 6 x + 5 x^{- 1}

f (x) = - x^{4} + 4 x^{3} + 7

f (x) = \frac{x}{( 1 + x ^{2} )}

y = 13 x^{4} + 4 x^{2} + 3

f (x) = \frac{2 x ^{2}}{x - 3}

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)})