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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

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

Minimum (- 1, 0)

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

f (x) = - x + 3 x + 5

f (x, y) = x^{3} + y^{3} + 3 x^{2} - 6 y^{2}

- 2 + 2 x - x^{2}

f (x) = \frac{x ^{3}}{3} + 8 x^{2} - 18 x + 7

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

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