What is the extreme f(x)=(x+4)/(sqrt(x^2+4)) ?

The extreme f(x)=(x+4)/(sqrt(x^2+4)) is Maximum(1,sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x+4)/(sqrt(x^2+4))

extreme points

f (x) = \frac{x + 4}{x ^{2} + 4}

Maximum (1, 5)

Solution steps

f^{'} (x) = \frac{- 4 x + 4}{( x ^{2} + 4 ) x ^{2} + 4}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

\frac{x + 4}{x ^{2} + 4} : 5

Maximum (1, 5)

f (x) = 6 - 5 x - x^{3}

f (x) = \frac{ln ( x )}{9 x}, 1 \leq x \leq 4

f (x) = \frac{x ^{2} - 2 x + 1}{x - 9}

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

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

What is the extreme f(x)=(x+4)/(sqrt(x^2+4)) ?
The extreme f(x)=(x+4)/(sqrt(x^2+4)) is Maximum(1,sqrt(5))