What is the extreme f(x)=x^4-72x+5 ?

The extreme f(x)=x^4-72x+5 is Minimum(\sqrt[3]{18},-54\sqrt[3]{18}+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^4-72x+5

extreme points

f (x) = x^{4} - 72 x + 5

Solution steps

f^{'} (x) = 4 x^{3} - 72

Find intervals:

Decreasing

Increasing

Plug

into

f (x, y) = 3 e^{x^{2} + y^{2}}

f (x, y) = (x - y) (2 x + y) (x + 1)

f (x y) = 2 x^{2} + 2 x y + y^{2} + 2 x - 3

2 x^{2} + 9

y = 2 x^{4} - 12 x^{2}

What is the extreme f(x)=x^4-72x+5 ?
The extreme f(x)=x^4-72x+5 is Minimum(\sqrt[3]{18},-54\sqrt[3]{18}+5)