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

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

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

extreme points

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)

\frac{d}{d M} (- M + \frac{1}{8} r a)

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

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

f (x, y) = 8 y^{2} - 8 x^{2} - 8 y + 3 x y

f (x) = 2 x^{3} - 24 x + 2

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