What is the extreme f(x)=(x^2-25)^{1/5} ?

The extreme f(x)=(x^2-25)^{1/5} is Minimum(0,-25^{1/5})

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)^{1/5}

extreme points

f (x) = (x^{2} - 25)^{\frac{1}{5}}

Minimum (0, - 2 5^{\frac{1}{5}})

Solution steps

f^{'} (x) = \frac{2 x}{5 ( x ^{2} - 25 ) ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 0

into

(x^{2} - 25)^{\frac{1}{5}} : - 2 5^{\frac{1}{5}}

Minimum (0, - 2 5^{\frac{1}{5}})

f (x) = 9 (x - 2)^{\frac{2}{3}}

f (x) = x^{2}

f (x) = \frac{1}{x - 5}

f (x) = 4^{x - 2}

What is the extreme f(x)=(x^2-25)^{1/5} ?
The extreme f(x)=(x^2-25)^{1/5} is Minimum(0,-25^{1/5})