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

The extreme f(x)=x^3-3x^2+12,-2<= x<= 4 is Minimum(-2,-8),Maximum(0,12),Minimum(2,8),Maximum(4,28)

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

extreme points

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

Minimum (- 2, - 8), Maximum (0, 12), Minimum (2, 8), Maximum (4, 28)

Solution steps

Find the critical points:

x = - 2, x = 0, x = 2, x = 4

Domain of

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 4

Plug the extreme point

x = - 2

into

x^{3} - 3 x^{2} + 12 \Rightarrow y = - 8

Minimum (- 2, - 8)

Plug the extreme point

x = 0

into

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

Maximum (0, 12)

Plug the extreme point

x = 2

into

x^{3} - 3 x^{2} + 12 \Rightarrow y = 8

Minimum (2, 8)

Plug the extreme point

x = 4

into

x^{3} - 3 x^{2} + 12 \Rightarrow y = 28

Maximum (4, 28)

Minimum (- 2, - 8), Maximum (0, 12), Minimum (2, 8), Maximum (4, 28)

f (x) = \frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x

f (x) = 6 x + 12 y - x^{2} - 3 y^{2}

f (x) = - 0.002 x^{2} + 4.4 x - 10

f (x) = x^{\frac{2}{5}} (x - 8)

f (x) = \frac{sin ( x )}{2 + cos ( x )}

What is the extreme f(x)=x^3-3x^2+12,-2<= x<= 4 ?
The extreme f(x)=x^3-3x^2+12,-2<= x<= 4 is Minimum(-2,-8),Maximum(0,12),Minimum(2,8),Maximum(4,28)