What is the extreme f(x)=(x^3-1)^4,-1<= x<= 2 ?

The extreme f(x)=(x^3-1)^4,-1<= x<= 2 is Maximum(-1,16),Saddle(0,1),Minimum(1,0),Maximum(2,2401)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x^3-1)^4,-1<= x<= 2

extreme points

f (x) = (x^{3} - 1)^{4}, - 1 \leq x \leq 2

Maximum (- 1, 16), Saddle (0, 1), Minimum (1, 0), Maximum (2, 2401)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1, x = 2

Domain of

(x^{3} - 1)^{4}, - 1 \leq x \leq 2 : - 1 \leq x \leq 2

Find intervals:

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = - 1

into

(x^{3} - 1)^{4} \Rightarrow y = 16

Maximum (- 1, 16)

Plug the extreme point

x = 0

into

(x^{3} - 1)^{4} \Rightarrow y = 1

Saddle (0, 1)

Plug the extreme point

x = 1

into

(x^{3} - 1)^{4} \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = 2

into

(x^{3} - 1)^{4} \Rightarrow y = 2401

Maximum (2, 2401)

Maximum (- 1, 16), Saddle (0, 1), Minimum (1, 0), Maximum (2, 2401)

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

f (x) = 8 x + 6 y - 1 = 0

10 x e^{- 2 x}

f (x) = \frac{e ^{x}}{x ^{3}}, x > 0

f (x, y) = ln (y - ∣ x ∣ - 1)

What is the extreme f(x)=(x^3-1)^4,-1<= x<= 2 ?
The extreme f(x)=(x^3-1)^4,-1<= x<= 2 is Maximum(-1,16),Saddle(0,1),Minimum(1,0),Maximum(2,2401)