What is the extreme f(x)=7x^3-3x^2+1 ?

The extreme f(x)=7x^3-3x^2+1 is Maximum(0,1),Minimum(2/7 , 45/49)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=7x^3-3x^2+1

extreme points

f (x) = 7 x^{3} - 3 x^{2} + 1

Maximum (0, 1), Minimum (\frac{2}{7}, \frac{45}{49})

Solution steps

f^{'} (x) = 21 x^{2} - 6 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{2}{7},

Increasing

: \frac{2}{7} < x < \infty

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = \frac{2}{7}

into

7 x^{3} - 3 x^{2} + 1 : \frac{45}{49}

Minimum (\frac{2}{7}, \frac{45}{49})

Maximum (0, 1), Minimum (\frac{2}{7}, \frac{45}{49})

- 7 cos (2 x)

y = 1. 1^{x}

1 + ln (x - 4)

(7, 3), (- 2, - 2)

f (x) = \frac{5 x}{4 x ^{2} - x - 14}

What is the extreme f(x)=7x^3-3x^2+1 ?
The extreme f(x)=7x^3-3x^2+1 is Maximum(0,1),Minimum(2/7 , 45/49)