What is the extreme f(x)=x^2-2x-9,0<= x<= 3 ?

The extreme f(x)=x^2-2x-9,0<= x<= 3 is Maximum(0,-9),Minimum(1,-10),Maximum(3,-6)

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extreme f(x)=x^2-2x-9,0<= x<= 3

extreme points

f (x) = x^{2} - 2 x - 9, 0 \leq x \leq 3

Maximum (0, - 9), Minimum (1, - 10), Maximum (3, - 6)

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

x^{2} - 2 x - 9, 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

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

Maximum (0, - 9)

Plug the extreme point

x = 1

into

x^{2} - 2 x - 9 \Rightarrow y = - 10

Minimum (1, - 10)

Plug the extreme point

x = 3

into

x^{2} - 2 x - 9 \Rightarrow y = - 6

Maximum (3, - 6)

Maximum (0, - 9), Minimum (1, - 10), Maximum (3, - 6)

f (x) = tan (\frac{π x}{32}), 0 \leq x \leq 8

3 x^{3} + 9 x^{2} + 27 x - 21

f (x, y) = 5 x ln (1 y - 9) + 5 y - 5

f (x) = x^{3} - 12 x^{2} + 21 x - 8

f (x) = x^{3} - 12 x + 5, - 6 \leq x \leq 6

What is the extreme f(x)=x^2-2x-9,0<= x<= 3 ?
The extreme f(x)=x^2-2x-9,0<= x<= 3 is Maximum(0,-9),Minimum(1,-10),Maximum(3,-6)