What is the extreme f(x)=x^2-4,-3<= x<= 2 ?

The extreme f(x)=x^2-4,-3<= x<= 2 is Maximum(-3,5),Minimum(0,-4),Maximum(2,0)

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extreme f(x)=x^2-4,-3<= x<= 2

extreme points

f (x) = x^{2} - 4, - 3 \leq x \leq 2

Maximum (- 3, 5), Minimum (0, - 4), Maximum (2, 0)

Solution steps

Find the critical points:

x = - 3, x = 0, x = 2

Domain of

x^{2} - 4, - 3 \leq x \leq 2 : - 3 \leq x \leq 2

Find intervals:

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 3

into

x^{2} - 4 \Rightarrow y = 5

Maximum (- 3, 5)

Plug the extreme point

x = 0

into

x^{2} - 4 \Rightarrow y = - 4

Minimum (0, - 4)

Plug the extreme point

x = 2

into

x^{2} - 4 \Rightarrow y = 0

Maximum (2, 0)

Maximum (- 3, 5), Minimum (0, - 4), Maximum (2, 0)

f (x) = (7.8 x^{2} - 225 x + 1709)

f (x) = - 2 x^{3} - 2 y^{3} + 6 x y + 10

8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1

f (x) = 3 x^{2} - 2 x + y^{2} - 4 y + 1

P (a, b) = 2 a + b

What is the extreme f(x)=x^2-4,-3<= x<= 2 ?
The extreme f(x)=x^2-4,-3<= x<= 2 is Maximum(-3,5),Minimum(0,-4),Maximum(2,0)