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

The extreme f(x)=2x^2-4x,0<= x<= 3 is Maximum(0,0),Minimum(1,-2),Maximum(3,6)

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

extreme points

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

Maximum (0, 0), Minimum (1, - 2), Maximum (3, 6)

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

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

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 3

Plug the extreme point

x = 0

into

2 x^{2} - 4 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

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

Minimum (1, - 2)

Plug the extreme point

x = 3

into

2 x^{2} - 4 x \Rightarrow y = 6

Maximum (3, 6)

Maximum (0, 0), Minimum (1, - 2), Maximum (3, 6)

x^{2} - 18 x + 79

f (x) = 3 x^{4} - 294 x^{2} + 3, - 8 \leq x \leq 8

f (x) = 2 x^{2} - 7, - 7 \leq x \leq 7

x^{2} - 18 x + 86

2 x^{2} - 4 x + 6

What is the extreme f(x)=2x^2-4x,0<= x<= 3 ?
The extreme f(x)=2x^2-4x,0<= x<= 3 is Maximum(0,0),Minimum(1,-2),Maximum(3,6)