What is the extreme f(x)=(1/12 x^3-9x+5) ?

The extreme f(x)=(1/12 x^3-9x+5) is Maximum(-6,41),Minimum(6,-31)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(1/12 x^3-9x+5)

extreme points

f (x) = (\frac{1}{12} x^{3} - 9 x + 5)

Maximum (- 6, 41), Minimum (6, - 31)

Solution steps

f^{'} (x) = \frac{x ^{2}}{4} - 9

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - 6

into

\frac{1}{12} x^{3} - 9 x + 5 : 41

Maximum (- 6, 41)

Plug

x = 6

into

\frac{1}{12} x^{3} - 9 x + 5 : - 31

Minimum (6, - 31)

Maximum (- 6, 41), Minimum (6, - 31)

f (x) = 2 x^{2} + 18 x + 16

y = 3 x^{2} = 12 x + 11

f (x) = x^{3} - 4 x^{2} - 3 x - 6

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

f (x) - \frac{1}{2} (x + 1)^{2} - \frac{3}{2}

What is the extreme f(x)=(1/12 x^3-9x+5) ?
The extreme f(x)=(1/12 x^3-9x+5) is Maximum(-6,41),Minimum(6,-31)