What is the extreme f(x)=2x^2-8x+11,0<= x<= 6 ?

The extreme f(x)=2x^2-8x+11,0<= x<= 6 is Maximum(0,11),Minimum(2,3),Maximum(6,35)

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extreme f(x)=2x^2-8x+11,0<= x<= 6

extreme points

f (x) = 2 x^{2} - 8 x + 11, 0 \leq x \leq 6

Maximum (0, 11), Minimum (2, 3), Maximum (6, 35)

Solution steps

Find the critical points:

x = 0, x = 2, x = 6

Domain of

2 x^{2} - 8 x + 11, 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 + 11 \Rightarrow y = 11

Maximum (0, 11)

Plug the extreme point

x = 2

into

2 x^{2} - 8 x + 11 \Rightarrow y = 3

Minimum (2, 3)

Plug the extreme point

x = 6

into

2 x^{2} - 8 x + 11 \Rightarrow y = 35

Maximum (6, 35)

Maximum (0, 11), Minimum (2, 3), Maximum (6, 35)

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

f (x) = x^{\frac{3}{2}} (x - 5)

f (x, y) = x^{2} - y^{2} - 10 x + 2 y + 8

f (x) = 9 - x^{2} es

- x^{2} + 4 x - 2

What is the extreme f(x)=2x^2-8x+11,0<= x<= 6 ?
The extreme f(x)=2x^2-8x+11,0<= x<= 6 is Maximum(0,11),Minimum(2,3),Maximum(6,35)