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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 0, x = 1

Domain of

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

Find intervals:

Decreasing

: 0 < x < 1

Plug the extreme point

x = 0

into

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

Maximum (0, 1)

Plug the extreme point

x = 1

into

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

Minimum (1, 0)

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

2 x^{2} + 12 x - 2

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

y = - x^{6} - 3 x^{3} + x + 1, (0, 1)

f (x, y) = x^{2} + 12 \cdot x \cdot y + 8 \cdot y^{2}

f (x) = - 5 x^{2} + 4 x - 6

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