What is the extreme f(x)=2x^5ln(x),(0,10) ?

The extreme f(x)=2x^5ln(x),(0,10) is Minimum(1/(\sqrt[5]{e)},-2/(5e))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=2x^5ln(x),(0,10)

extreme points

f (x) = 2 x^{5} ln (x), (0, 10)

Solution steps

Find the critical points:

Domain of

2 x^{5} ln (x), (0, 10) : 0 < x < 10

Find intervals:

Decreasing

Increasing

Plug the extreme point

into

2 x^{5} ln (x) \Rightarrow y = - \frac{2}{5 e}

f (x) = \frac{2 x - 7}{x}

8 x^{2} - 8 x + 3

p (x) = 9 x^{4} + 9 x^{2} - 18

f (x) = 2 x^{3} + 6 x^{2} - 48 x + 6 [4.6]

f (x) = 3 sin (x), - π \leq x \leq 2 π

What is the extreme f(x)=2x^5ln(x),(0,10) ?
The extreme f(x)=2x^5ln(x),(0,10) is Minimum(1/(\sqrt[5]{e)},-2/(5e))