What is the extreme f(x)=((x^2+25))/(2x) ?

The extreme f(x)=((x^2+25))/(2x) is Maximum(-5,-5),Minimum(5,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=((x^2+25))/(2x)

extreme points

f (x) = \frac{( x ^{2} + 25 )}{2 x}

Maximum (- 5, - 5), Minimum (5, 5)

Solution steps

f^{'} (x) = \frac{x ^{2} - 25}{2 x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 5

into

\frac{x ^{2} + 25}{2 x} : - 5

Maximum (- 5, - 5)

Plug

x = 5

into

\frac{x ^{2} + 25}{2 x} : 5

Minimum (5, 5)

Maximum (- 5, - 5), Minimum (5, 5)

x - 3 = 0

f (x) = \frac{2 x ^{2} - 9 x - 5}{x ^{2} - 16}

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

- x + 3

\frac{- 4 - 3 x}{7 x - 5}

What is the extreme f(x)=((x^2+25))/(2x) ?
The extreme f(x)=((x^2+25))/(2x) is Maximum(-5,-5),Minimum(5,5)