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

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

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

extreme points

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

Maximum (- 2, 4), Minimum (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

x = - 2, x = 0, x = 1

Domain of

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

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

x^{2} \Rightarrow y = 4

Maximum (- 2, 4)

Plug the extreme point

x = 0

into

x^{2} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

x^{2} \Rightarrow y = 1

Maximum (1, 1)

Maximum (- 2, 4), Minimum (0, 0), Maximum (1, 1)

f (x) = \frac{x ^{2} - 27}{x - 6}

f (x) = - 2 x^{3} + 3 x^{2} + 12 x + 2

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

f (x, y) = ln (\frac{1}{x + y})

f (x, y) = 4 x + 6 y

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