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

The extreme f(x)=3x^3+3/2 x^2-2x,-1<= x<= 1 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=3x^3+3/2 x^2-2x,-1<= x<= 1

extreme points

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

Minimum (- 1, \frac{1}{2}), Maximum (- \frac{2}{3}, \frac{10}{9}), Minimum (\frac{1}{3}, - \frac{7}{18}), Maximum (1, \frac{5}{2})

Solution steps

Find the critical points:

x = - 1, x = - \frac{2}{3}, x = \frac{1}{3}, x = 1

Domain of

3 x^{3} + \frac{3}{2} x^{2} - 2 x, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Increasing

: - 1 < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < \frac{1}{3},

Increasing

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

Plug the extreme point

x = - 1

into

3 x^{3} + \frac{3}{2} x^{2} - 2 x \Rightarrow y = \frac{1}{2}

Minimum (- 1, \frac{1}{2})

Plug the extreme point

x = - \frac{2}{3}

into

3 x^{3} + \frac{3}{2} x^{2} - 2 x \Rightarrow y = \frac{10}{9}

Maximum (- \frac{2}{3}, \frac{10}{9})

Plug the extreme point

x = \frac{1}{3}

into

3 x^{3} + \frac{3}{2} x^{2} - 2 x \Rightarrow y = - \frac{7}{18}

Minimum (\frac{1}{3}, - \frac{7}{18})

Plug the extreme point

x = 1

into

3 x^{3} + \frac{3}{2} x^{2} - 2 x \Rightarrow y = \frac{5}{2}

Maximum (1, \frac{5}{2})

Minimum (- 1, \frac{1}{2}), Maximum (- \frac{2}{3}, \frac{10}{9}), Minimum (\frac{1}{3}, - \frac{7}{18}), Maximum (1, \frac{5}{2})

2 - \frac{10 x}{x ^{2} + 25}

f (x) = 7 + 5 x - 5 x^{2}

f (x) = e^{- x} (x^{2} - 1)

h (x) = 2 (x - 5)^{3}

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

What is the extreme f(x)=3x^3+3/2 x^2-2x,-1<= x<= 1 ?
The extreme f(x)=3x^3+3/2 x^2-2x,-1<= x<= 1 is