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 Minimum(0,-1),Maximum(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

Minimum (0, - 1), Maximum (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:

Increasing

: 0 < x < 1

Plug the extreme point

x = 0

into

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

Minimum (0, - 1)

Plug the extreme point

x = 1

into

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

Maximum (1, 0)

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

f (x) = x^{2} + x y + \frac{1}{2} y^{2} - 3 x + y

x^{2} + \frac{400}{x}

0.05 x + 20 + \frac{125}{x}

f (x) = \frac{9 ( t - 2 ) ^{2}}{t ^{2}}

x y - 5 x - 5 y - x^{2} - y^{2}

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