What is the extreme f(x)=-2x^3+15x^2-24x,0<= x<= 5 ?

The extreme f(x)=-2x^3+15x^2-24x,0<= x<= 5 is Maximum(0,0),Minimum(1,-11),Maximum(4,16),Minimum(5,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=-2x^3+15x^2-24x,0<= x<= 5

extreme points

f (x) = - 2 x^{3} + 15 x^{2} - 24 x, 0 \leq x \leq 5

Maximum (0, 0), Minimum (1, - 11), Maximum (4, 16), Minimum (5, 5)

Solution steps

Find the critical points:

x = 0, x = 1, x = 4, x = 5

Domain of

- 2 x^{3} + 15 x^{2} - 24 x, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 4,

Decreasing

: 4 < x < 5

Plug the extreme point

x = 0

into

- 2 x^{3} + 15 x^{2} - 24 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

- 2 x^{3} + 15 x^{2} - 24 x \Rightarrow y = - 11

Minimum (1, - 11)

Plug the extreme point

x = 4

into

- 2 x^{3} + 15 x^{2} - 24 x \Rightarrow y = 16

Maximum (4, 16)

Plug the extreme point

x = 5

into

- 2 x^{3} + 15 x^{2} - 24 x \Rightarrow y = 5

Minimum (5, 5)

Maximum (0, 0), Minimum (1, - 11), Maximum (4, 16), Minimum (5, 5)

\frac{e ^{x}}{6 + e ^{x}}

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

f (x, y) = 4 x^{2} + 2 y^{2} + 6 x + 8 y + 2

y = \frac{2 x}{x ^{2} - 1}

x^{\frac{2}{3}} (x - 5), - 5 \leq x \leq 5

What is the extreme f(x)=-2x^3+15x^2-24x,0<= x<= 5 ?
The extreme f(x)=-2x^3+15x^2-24x,0<= x<= 5 is Maximum(0,0),Minimum(1,-11),Maximum(4,16),Minimum(5,5)