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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = - 1, x = 0

Domain of

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

Find intervals:

Decreasing

: - 1 < x < 0

Plug the extreme point

x = - 1

into

x^{2} \Rightarrow y = 1

Maximum (- 1, 1)

Plug the extreme point

x = 0

into

x^{2} \Rightarrow y = 0

Minimum (0, 0)

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

3 x^{2} - 5 x + 2

f (x) = x^{4} - 4 x^{3} + 16 x + 4

f (x) = x^{7} - 7 x

f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

y = 2 x^{2} - 8 x + 8

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