What is the extreme f(x)=(x+6)/(sqrt(x^2+6)) ?

The extreme f(x)=(x+6)/(sqrt(x^2+6)) is Maximum(1,sqrt(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x+6)/(sqrt(x^2+6))

extreme points

f (x) = \frac{x + 6}{x ^{2} + 6}

Maximum (1, 7)

Solution steps

f^{'} (x) = \frac{- 6 x + 6}{( x ^{2} + 6 ) x ^{2} + 6}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

\frac{x + 6}{x ^{2} + 6} : 7

Maximum (1, 7)

ln (x^{2} + 49)

f (x, y) = 9 - 2 x^{2} - y^{2}

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

f (x) = - x^{3} - 9 x^{2} + 165 x + 1300

f (x, y) = (x^{2} - 1) (y^{2} - 1)

What is the extreme f(x)=(x+6)/(sqrt(x^2+6)) ?
The extreme f(x)=(x+6)/(sqrt(x^2+6)) is Maximum(1,sqrt(7))