What is the extreme y= 1/(\sqrt[3]{1-x^2)} ?

The extreme y= 1/(\sqrt[3]{1-x^2)} is Minimum(0,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme y= 1/(\sqrt[3]{1-x^2)}

extreme points

y = \frac{1}{3 1 - x ^{2}}

Minimum (0, 1)

Solution steps

f^{'} (x) = \frac{2 x}{3 ( 1 - x ^{2} ) ^{\frac{4}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

\frac{1}{3 1 - x ^{2}} : 1

Minimum (0, 1)

f (x) = 2 x^{2} - 8 x + 2, 0 \leq x \leq 5

f (x) = 10 x + 13 x^{\frac{10}{13}}

f (x) = x^{3} e^{- 4 x}

f (x, y) = e^{- x^{2} - y^{2}} \cdot (x^{2} + 2 y^{2})

f (x) = 7 x^{2} - 28 x + 3, - 5 \leq x \leq 4

What is the extreme y= 1/(\sqrt[3]{1-x^2)} ?
The extreme y= 1/(\sqrt[3]{1-x^2)} is Minimum(0,1)