What is the extreme f(x)=x^2-8x-9,1<= x<= 6 ?

The extreme f(x)=x^2-8x-9,1<= x<= 6 is Maximum(1,-16),Minimum(4,-25),Maximum(6,-21)

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extreme f(x)=x^2-8x-9,1<= x<= 6

extreme points

f (x) = x^{2} - 8 x - 9, 1 \leq x \leq 6

Maximum (1, - 16), Minimum (4, - 25), Maximum (6, - 21)

Solution steps

Find the critical points:

x = 1, x = 4, x = 6

Domain of

x^{2} - 8 x - 9, 1 \leq x \leq 6 : 1 \leq x \leq 6

Find intervals:

Decreasing

: 1 < x < 4,

Increasing

: 4 < x < 6

Plug the extreme point

x = 1

into

x^{2} - 8 x - 9 \Rightarrow y = - 16

Maximum (1, - 16)

Plug the extreme point

x = 4

into

x^{2} - 8 x - 9 \Rightarrow y = - 25

Minimum (4, - 25)

Plug the extreme point

x = 6

into

x^{2} - 8 x - 9 \Rightarrow y = - 21

Maximum (6, - 21)

Maximum (1, - 16), Minimum (4, - 25), Maximum (6, - 21)

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

f (x) = x^{2} - 8 x + 19

f (x) = - 0.02 x^{2} + 4 x - 40

f (x, y) = 2 x^{3} + y^{3} + 3 x^{2} - 3 y^{2} - 16

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

What is the extreme f(x)=x^2-8x-9,1<= x<= 6 ?
The extreme f(x)=x^2-8x-9,1<= x<= 6 is Maximum(1,-16),Minimum(4,-25),Maximum(6,-21)