What is the extreme f(x)=39+6x+(54)/x ,0<x<27 ?

The extreme f(x)=39+6x+(54)/x ,0<x<27 is Minimum(3,75)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=39+6x+(54)/x ,0<x<27

extreme points

f (x) = 39 + 6 x + \frac{54}{x}, 0 < x < 27

Minimum (3, 75)

Solution steps

Find the critical points:

x = 3

Domain of

39 + 6 x + \frac{54}{x}, 0 < x < 27 : 0 < x < 27

Find intervals:

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < 27

Plug the extreme point

x = 3

into

39 + 6 x + \frac{54}{x} \Rightarrow y = 75

Minimum (3, 75)

x + 12 x^{- 1} + 18 x^{- 2}

p (a, b) = \frac{3 ( \frac{600 ( a - b )}{a + b} ) + 1800}{a}

f (x) = x^{3} - 2 x^{2} - 15 x

y = x^{3} - \frac{3}{2} x^{2}

f (x) = 5 x^{2} - 8 x + 3

What is the extreme f(x)=39+6x+(54)/x ,0<x<27 ?
The extreme f(x)=39+6x+(54)/x ,0<x<27 is Minimum(3,75)