What is the extreme f(x)=sqrt(x)(x+4),x>= 0 ?

The extreme f(x)=sqrt(x)(x+4),x>= 0 is Minimum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=sqrt(x)(x+4),x>= 0

extreme points

f (x) = x (x + 4), x \geq 0

Minimum (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

x (x + 4), x \geq 0 : x \geq 0

Find intervals:

Increasing

: 0 < x < \infty

Plug the extreme point

x = 0

into

x (x + 4) \Rightarrow y = 0

Minimum (0, 0)

f (x) = 190 + 8 x^{3} + x^{4}

y = 4 - 6 x + x^{2}

f (x) = 0.002 x^{2} + 3.4 x - 20

f (x, y) = x^{3} + 2 y^{2} - 27 x - 8 y - 4

f (x) = x^{\frac{1}{3}} + 9

What is the extreme f(x)=sqrt(x)(x+4),x>= 0 ?
The extreme f(x)=sqrt(x)(x+4),x>= 0 is Minimum(0,0)