What is the extreme f(x)=2x^2-8x,0<= x<= 6 ?

The extreme f(x)=2x^2-8x,0<= x<= 6 is Maximum(0,0),Minimum(2,-8),Maximum(6,24)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=2x^2-8x,0<= x<= 6

extreme points

f (x) = 2 x^{2} - 8 x, 0 \leq x \leq 6

Maximum (0, 0), Minimum (2, - 8), Maximum (6, 24)

Solution steps

Find the critical points:

x = 0, x = 2, x = 6

Domain of

2 x^{2} - 8 x, 0 \leq x \leq 6 : 0 \leq x \leq 6

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 6

Plug the extreme point

x = 0

into

2 x^{2} - 8 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 2

into

2 x^{2} - 8 x \Rightarrow y = - 8

Minimum (2, - 8)

Plug the extreme point

x = 6

into

2 x^{2} - 8 x \Rightarrow y = 24

Maximum (6, 24)

Maximum (0, 0), Minimum (2, - 8), Maximum (6, 24)

What is the extreme f(x)=2x^2-8x,0<= x<= 6 ?
The extreme f(x)=2x^2-8x,0<= x<= 6 is Maximum(0,0),Minimum(2,-8),Maximum(6,24)