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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = - 1, x = 0, x = 2

Domain of

x^{2} - 1, - 1 \leq x \leq 2 : - 1 \leq x \leq 2

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 1

into

x^{2} - 1 \Rightarrow y = 0

Maximum (- 1, 0)

Plug the extreme point

x = 0

into

x^{2} - 1 \Rightarrow y = - 1

Minimum (0, - 1)

Plug the extreme point

x = 2

into

x^{2} - 1 \Rightarrow y = 3

Maximum (2, 3)

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

2 cos (x)

f (x) = 6 + 2

8 x + 4

\frac{9}{x ^{2} - 16}

(- 2, - 1), (- 8, 6)

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