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

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

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

extreme points

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

Minimum (0, 3), Maximum (1, 4), Minimum (2, 3)

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2

Plug the extreme point

x = 0

into

x^{2} - 2 x - 3 \Rightarrow y = 3

Minimum (0, 3)

Plug the extreme point

x = 1

into

x^{2} - 2 x - 3 \Rightarrow y = 4

Maximum (1, 4)

Plug the extreme point

x = 2

into

x^{2} - 2 x - 3 \Rightarrow y = 3

Minimum (2, 3)

Minimum (0, 3), Maximum (1, 4), Minimum (2, 3)

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