What is the extreme f(x)=(x+1)/(x^2+3) ?

The extreme f(x)=(x+1)/(x^2+3) is Minimum(-3,-1/6),Maximum(1, 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+1)/(x^2+3)

extreme points

f (x) = \frac{x + 1}{x ^{2} + 3}

Minimum (- 3, - \frac{1}{6}), Maximum (1, \frac{1}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 3

into

\frac{x + 1}{x ^{2} + 3} : - \frac{1}{6}

Minimum (- 3, - \frac{1}{6})

Plug

x = 1

into

\frac{x + 1}{x ^{2} + 3} : \frac{1}{2}

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

Minimum (- 3, - \frac{1}{6}), Maximum (1, \frac{1}{2})

f (x, y) = e^{5 x y}

f (x, y) = 2 x - y

9 x^{2} ln (x)

f (x) = x^{3} + \frac{3}{2} y^{2} - 3 x y - 124

f (x, y) = 2 x + 2 y

What is the extreme f(x)=(x+1)/(x^2+3) ?
The extreme f(x)=(x+1)/(x^2+3) is Minimum(-3,-1/6),Maximum(1, 1/2)