What is the extreme f(x)=-x^2+12,-3<= x<= 4 ?

The extreme f(x)=-x^2+12,-3<= x<= 4 is Minimum(-3,3),Maximum(0,12),Minimum(4,-4)

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extreme f(x)=-x^2+12,-3<= x<= 4

extreme points

f (x) = - x^{2} + 12, - 3 \leq x \leq 4

Minimum (- 3, 3), Maximum (0, 12), Minimum (4, - 4)

Solution steps

Find the critical points:

x = - 3, x = 0, x = 4

Domain of

- x^{2} + 12, - 3 \leq x \leq 4 : - 3 \leq x \leq 4

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 4

Plug the extreme point

x = - 3

into

- x^{2} + 12 \Rightarrow y = 3

Minimum (- 3, 3)

Plug the extreme point

x = 0

into

- x^{2} + 12 \Rightarrow y = 12

Maximum (0, 12)

Plug the extreme point

x = 4

into

- x^{2} + 12 \Rightarrow y = - 4

Minimum (4, - 4)

Minimum (- 3, 3), Maximum (0, 12), Minimum (4, - 4)

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 2 x + 10 y

2 x^{3} - 24 x + 1

f (x) = x^{7} e^{2 x} + 2

f (x) = x^{3} + 12 x y + y^{4}

f (x) = 3 x - 5 ln (4 x)

What is the extreme f(x)=-x^2+12,-3<= x<= 4 ?
The extreme f(x)=-x^2+12,-3<= x<= 4 is Minimum(-3,3),Maximum(0,12),Minimum(4,-4)