What is the extreme f(x)=(9x)/(sqrt(x-4)) ?

The extreme f(x)=(9x)/(sqrt(x-4)) is Minimum(8,36)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(9x)/(sqrt(x-4))

extreme points

f (x) = \frac{9 x}{x - 4}

Minimum (8, 36)

Solution steps

f^{'} (x) = \frac{9 ( x - 8 )}{2 ( x - 4 ) x - 4}

Find intervals:

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 8

into

\frac{9 x}{x - 4} : 36

Minimum (8, 36)

f (x) = x e^{\frac{- x ^{2}}{32}}

2400 = 30 x \cdot 10 y

f (- 1.2) = 3 p^{3} + 4 q^{3} - 2 p^{2} q - 4 p - 5 q

f (x) = 8 e x - y - 5 x y + 6 y^{2}

f (x) = - \frac{1}{50} x^{2} + 1.85 x - 8.5

What is the extreme f(x)=(9x)/(sqrt(x-4)) ?
The extreme f(x)=(9x)/(sqrt(x-4)) is Minimum(8,36)