What is the extreme y=-x^2+1,-2<= x<= 1 ?

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

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extreme y=-x^2+1,-2<= x<= 1

extreme points

y = - x^{2} + 1, - 2 \leq x \leq 1

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

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

Minimum (- 2, - 3)

Plug the extreme point

x = 0

into

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

Maximum (0, 1)

Plug the extreme point

x = 1

into

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

Minimum (1, 0)

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

f (x) = x^{3} - 6 x^{2} - 15 x

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

f (x) = 4 x + \frac{9}{x}

f (x) = x^{3} + 3 x^{2} + 3 x

f (x) = \frac{1}{4} x^{4} - x^{3} - 4 x + 2

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