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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Maximum (- 2, - 1), Minimum (2, 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x ^{2} + 4}{4 x} : - 1

Maximum (- 2, - 1)

Plug

x = 2

into

\frac{x ^{2} + 4}{4 x} : 1

Minimum (2, 1)

Maximum (- 2, - 1), Minimum (2, 1)

f (x) = - 6 x^{2}

\frac{x + 3}{x - 9}

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

y = \frac{x}{x ^{2} - 4}

\frac{x ^{2}}{x ^{2} - 9}

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