What is the extreme f(x)=2x^2-8x+2,0<= x<= 5 ?

The extreme f(x)=2x^2-8x+2,0<= x<= 5 is Maximum(0,2),Minimum(2,-6),Maximum(5,12)

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extreme f(x)=2x^2-8x+2,0<= x<= 5

extreme

Maximum (0, 2), Minimum (2, - 6), Maximum (5, 12)

Solution steps

Find the critical points:

x = 0, x = 2, x = 5

Domain of

2 x^{2} - 8 x + 2, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 5

Plug the extreme point

x = 0

into

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

Maximum (0, 2)

Plug the extreme point

x = 2

into

2 x^{2} - 8 x + 2 \Rightarrow y = - 6

Minimum (2, - 6)

Plug the extreme point

x = 5

into

2 x^{2} - 8 x + 2 \Rightarrow y = 12

Maximum (5, 12)

Maximum (0, 2), Minimum (2, - 6), Maximum (5, 12)

What is the extreme f(x)=2x^2-8x+2,0<= x<= 5 ?
The extreme f(x)=2x^2-8x+2,0<= x<= 5 is Maximum(0,2),Minimum(2,-6),Maximum(5,12)