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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: - 1 < x < 2,

Decreasing

: 2 < x < 2

Plug the extreme point

x = - 1

into

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

Minimum (- 1, - 3)

Plug the extreme point

x = 2

into

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

Maximum (2, 2)

Plug the extreme point

x = 2

into

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

Minimum (2, 0)

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

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

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

f (x) = \frac{1}{x + 2}

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 6 y^{2} - 9 x

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

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