What is the extreme f(x)=3(x-4)(x+7)^2 ?

The extreme f(x)=3(x-4)(x+7)^2 is Maximum(-7,0),Minimum(1/3 ,-5324/9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3(x-4)(x+7)^2

extreme points

f (x) = 3 (x - 4) (x + 7)^{2}

Maximum (- 7, 0), Minimum (\frac{1}{3}, - \frac{5324}{9})

Solution steps

f^{'} (x) = 3 (3 x^{2} + 20 x - 7)

Find intervals:

Increasing

: - \infty < x < - 7,

Decreasing

: - 7 < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < \infty

Plug

x = - 7

into

3 (x - 4) (x + 7)^{2} : 0

Maximum (- 7, 0)

Plug

x = \frac{1}{3}

into

3 (x - 4) (x + 7)^{2} : - \frac{5324}{9}

Minimum (\frac{1}{3}, - \frac{5324}{9})

Maximum (- 7, 0), Minimum (\frac{1}{3}, - \frac{5324}{9})

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

x^{3} - 3 x^{2} + 3 x + 1

100 + \frac{1}{2} x + \frac{1800}{x}

- cos (x)

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

What is the extreme f(x)=3(x-4)(x+7)^2 ?
The extreme f(x)=3(x-4)(x+7)^2 is Maximum(-7,0),Minimum(1/3 ,-5324/9)