What is the extreme f(x)=4x^2-16x+4,0<= x<= 3 ?

The extreme f(x)=4x^2-16x+4,0<= x<= 3 is Maximum(0,4),Minimum(2,-12),Maximum(3,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=4x^2-16x+4,0<= x<= 3

extreme points

f (x) = 4 x^{2} - 16 x + 4, 0 \leq x \leq 3

Maximum (0, 4), Minimum (2, - 12), Maximum (3, - 8)

Solution steps

Find the critical points:

x = 0, x = 2, x = 3

Domain of

4 x^{2} - 16 x + 4, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = 0

into

4 x^{2} - 16 x + 4 \Rightarrow y = 4

Maximum (0, 4)

Plug the extreme point

x = 2

into

4 x^{2} - 16 x + 4 \Rightarrow y = - 12

Minimum (2, - 12)

Plug the extreme point

x = 3

into

4 x^{2} - 16 x + 4 \Rightarrow y = - 8

Maximum (3, - 8)

Maximum (0, 4), Minimum (2, - 12), Maximum (3, - 8)

f (x, y) = 2 x^{2} + 4 x + 2 y^{2} + 5 y

f (x) = x^{3} - 2^{2}

- x^{3} + 5 x^{2} + 8 x + 9

- 2 x^{2} + 4 x - 8

- 2 x^{2} + 4 x - 3

What is the extreme f(x)=4x^2-16x+4,0<= x<= 3 ?
The extreme f(x)=4x^2-16x+4,0<= x<= 3 is Maximum(0,4),Minimum(2,-12),Maximum(3,-8)