What is the extreme f(x)=sqrt(-x^2+4),-1<= x<= 2 ?

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

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extreme f(x)=sqrt(-x^2+4),-1<= x<= 2

extreme points

f (x) = - x^{2} + 4, - 1 \leq x \leq 2

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

Solution steps

Find the critical points:

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

Domain of

- x^{2} + 4, - 1 \leq x \leq 2 : - 1 \leq x \leq 2

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2

Plug the extreme point

x = - 1

into

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

Minimum (- 1, 3)

Plug the extreme point

x = 0

into

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

Maximum (0, 2)

Plug the extreme point

x = 2

into

- x^{2} + 4 \Rightarrow y = 0

Minimum (2, 0)

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

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, 0 \leq x \leq 1

5 x^{3} - 30 x^{2} + 45 x

f (x) = - x + 2 cos (x), 0 \leq x \leq 2 π

+ (x - 1)^{2}

f (x) = 2 x^{\frac{5}{3}} - 3 x^{\frac{2}{3}}

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