What is the extreme f(x)=x^3+6x^2-63x+12,-5<= x<= 4 ?

The extreme f(x)=x^3+6x^2-63x+12,-5<= x<= 4 is Maximum(-5,352),Minimum(3,-96),Maximum(4,-80)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x^3+6x^2-63x+12,-5<= x<= 4

extreme points

f (x) = x^{3} + 6 x^{2} - 63 x + 12, - 5 \leq x \leq 4

Maximum (- 5, 352), Minimum (3, - 96), Maximum (4, - 80)

Solution steps

Find the critical points:

x = - 5, x = 3, x = 4

Domain of

x^{3} + 6 x^{2} - 63 x + 12, - 5 \leq x \leq 4 : - 5 \leq x \leq 4

Find intervals:

Decreasing

: - 5 < x < 3,

Increasing

: 3 < x < 4

Plug the extreme point

x = - 5

into

x^{3} + 6 x^{2} - 63 x + 12 \Rightarrow y = 352

Maximum (- 5, 352)

Plug the extreme point

x = 3

into

x^{3} + 6 x^{2} - 63 x + 12 \Rightarrow y = - 96

Minimum (3, - 96)

Plug the extreme point

x = 4

into

x^{3} + 6 x^{2} - 63 x + 12 \Rightarrow y = - 80

Maximum (4, - 80)

Maximum (- 5, 352), Minimum (3, - 96), Maximum (4, - 80)

9 sin (x) + 9 cos (x), 0 \leq x \leq 2 π

f (x) = sin (2 θ)

f (x) = - 0.001 x^{2} + 4.2 x - 10

f (x, y) = x y - x^{2} - y + 1

f (x, y) = x^{2} + x y + y^{2} + 6 x - 3 y + 4

What is the extreme f(x)=x^3+6x^2-63x+12,-5<= x<= 4 ?
The extreme f(x)=x^3+6x^2-63x+12,-5<= x<= 4 is Maximum(-5,352),Minimum(3,-96),Maximum(4,-80)