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

The extreme f(x)=x^3-x^2-12+2,0<= x<= 4 is Maximum(0,-10),Minimum(2/3 ,-4/27-10),Maximum(4,38)

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

extreme

Maximum (0, - 10), Minimum (\frac{2}{3}, - \frac{4}{27} - 10), Maximum (4, 38)

Solution steps

Find the critical points:

x = 0, x = \frac{2}{3}, x = 4

Domain of

x^{3} - x^{2} - 12 + 2, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Decreasing

: 0 < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < 4

Plug the extreme point

x = 0

into

x^{3} - x^{2} - 12 + 2 \Rightarrow y = - 10

Maximum (0, - 10)

Plug the extreme point

x = \frac{2}{3}

into

x^{3} - x^{2} - 12 + 2 \Rightarrow y = - \frac{4}{27} - 10

Minimum (\frac{2}{3}, - \frac{4}{27} - 10)

Plug the extreme point

x = 4

into

x^{3} - x^{2} - 12 + 2 \Rightarrow y = 38

Maximum (4, 38)

Maximum (0, - 10), Minimum (\frac{2}{3}, - \frac{4}{27} - 10), Maximum (4, 38)

What is the extreme f(x)=x^3-x^2-12+2,0<= x<= 4 ?
The extreme f(x)=x^3-x^2-12+2,0<= x<= 4 is Maximum(0,-10),Minimum(2/3 ,-4/27-10),Maximum(4,38)