What is the extreme f(x)=x^4+4x^3+15,-4<= x<= 1 ?

The extreme f(x)=x^4+4x^3+15,-4<= x<= 1 is Maximum(-4,15),Minimum(-3,-12),Saddle(0,15),Maximum(1,20)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x^4+4x^3+15,-4<= x<= 1

extreme points

f (x) = x^{4} + 4 x^{3} + 15, - 4 \leq x \leq 1

Maximum (- 4, 15), Minimum (- 3, - 12), Saddle (0, 15), Maximum (1, 20)

Solution steps

Find the critical points:

x = - 4, x = - 3, x = 0, x = 1

Domain of

x^{4} + 4 x^{3} + 15, - 4 \leq x \leq 1 : - 4 \leq x \leq 1

Find intervals:

Decreasing

: - 4 < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 4

into

x^{4} + 4 x^{3} + 15 \Rightarrow y = 15

Maximum (- 4, 15)

Plug the extreme point

x = - 3

into

x^{4} + 4 x^{3} + 15 \Rightarrow y = - 12

Minimum (- 3, - 12)

Plug the extreme point

x = 0

into

x^{4} + 4 x^{3} + 15 \Rightarrow y = 15

Saddle (0, 15)

Plug the extreme point

x = 1

into

x^{4} + 4 x^{3} + 15 \Rightarrow y = 20

Maximum (1, 20)

Maximum (- 4, 15), Minimum (- 3, - 12), Saddle (0, 15), Maximum (1, 20)

f (x) = cos (x) [- \frac{π}{4}, \frac{π}{4}]

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

f (x) = \frac{12}{6 - x} - \frac{x}{3} + 8

y = - 2 x^{2} + 12 x - 8

f (x) = x^{3} - 3 x^{2} - 9 x + 6

What is the extreme f(x)=x^4+4x^3+15,-4<= x<= 1 ?
The extreme f(x)=x^4+4x^3+15,-4<= x<= 1 is Maximum(-4,15),Minimum(-3,-12),Saddle(0,15),Maximum(1,20)