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

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

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extreme f(x)=sqrt(1-x^2),-1<= x<= 0

extreme points

f (x) = 1 - x^{2}, - 1 \leq x \leq 0

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

Solution steps

Find the critical points:

x = - 1, x = 0

Domain of

1 - x^{2}, - 1 \leq x \leq 0 : - 1 \leq x \leq 0

Find intervals:

Increasing

: - 1 < x < 0

Plug the extreme point

x = - 1

into

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

Minimum (- 1, 0)

Plug the extreme point

x = 0

into

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

Maximum (0, 1)

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

f (\frac{x}{y}) = x y + y x

f (x) = 2 x^{5} - 3 x^{3} + 8 x - 10

f (x) = x^{2} + 3 (x) (\frac{500}{x ^{2}})

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

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

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