What is the extreme y= x/(x^2+49) ?

The extreme y= x/(x^2+49) is Minimum(-7,-1/14),Maximum(7, 1/14)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme y= x/(x^2+49)

extreme points

y = \frac{x}{x ^{2} + 49}

Minimum (- 7, - \frac{1}{14}), Maximum (7, \frac{1}{14})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 49}{( x ^{2} + 49 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 7,

Increasing

: - 7 < x < 7,

Decreasing

: 7 < x < \infty

Plug

x = - 7

into

\frac{x}{x ^{2} + 49} : - \frac{1}{14}

Minimum (- 7, - \frac{1}{14})

Plug

x = 7

into

\frac{x}{x ^{2} + 49} : \frac{1}{14}

Maximum (7, \frac{1}{14})

Minimum (- 7, - \frac{1}{14}), Maximum (7, \frac{1}{14})

(x - 8) ln (x - 8)

f (x) = 12 + 2 x^{2} - x^{4}

f (x) = - 15 + 7 x - x^{2}

f (x) = \frac{0.002 x ^{2} + 24}{x}

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

What is the extreme y= x/(x^2+49) ?
The extreme y= x/(x^2+49) is Minimum(-7,-1/14),Maximum(7, 1/14)