What is the extreme f(x)=15x^4-4x^3 ?

The extreme f(x)=15x^4-4x^3 is Saddle(0,0),Minimum(1/5 ,-1/125)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=15x^4-4x^3

extreme points

f (x) = 15 x^{4} - 4 x^{3}

Saddle (0, 0), Minimum (\frac{1}{5}, - \frac{1}{125})

Solution steps

f^{'} (x) = 60 x^{3} - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{5},

Increasing

: \frac{1}{5} < x < \infty

Plug

x = 0

into

15 x^{4} - 4 x^{3} : 0

Saddle (0, 0)

Plug

x = \frac{1}{5}

into

15 x^{4} - 4 x^{3} : - \frac{1}{125}

Minimum (\frac{1}{5}, - \frac{1}{125})

Saddle (0, 0), Minimum (\frac{1}{5}, - \frac{1}{125})

f (x) = x^{3} - \frac{5}{2} x^{2} - 2 x

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

f (x, y) = x^{3} - 12 x - y^{2} - 8 y + 18

f (x) = 45 x + y

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 7

What is the extreme f(x)=15x^4-4x^3 ?
The extreme f(x)=15x^4-4x^3 is Saddle(0,0),Minimum(1/5 ,-1/125)