What is the extreme f(x)=x^3+3x^2-24x,0<= x<= 3 ?

The extreme f(x)=x^3+3x^2-24x,0<= x<= 3 is Maximum(0,0),Minimum(2,-28),Maximum(3,-18)

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extreme f(x)=x^3+3x^2-24x,0<= x<= 3

extreme points

f (x) = x^{3} + 3 x^{2} - 24 x, 0 \leq x \leq 3

Maximum (0, 0), Minimum (2, - 28), Maximum (3, - 18)

Solution steps

Find the critical points:

x = 0, x = 2, x = 3

Domain of

x^{3} + 3 x^{2} - 24 x, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = 0

into

x^{3} + 3 x^{2} - 24 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 2

into

x^{3} + 3 x^{2} - 24 x \Rightarrow y = - 28

Minimum (2, - 28)

Plug the extreme point

x = 3

into

x^{3} + 3 x^{2} - 24 x \Rightarrow y = - 18

Maximum (3, - 18)

Maximum (0, 0), Minimum (2, - 28), Maximum (3, - 18)

f (x) = 2 cos^{2} (x), (0, π)

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

f (x) = 1.5 x^{2} + 45 x + 15000

f (x) = x^{3} - 8 x^{2} + 3

f (x) = x^{3} - 12 x + 18

What is the extreme f(x)=x^3+3x^2-24x,0<= x<= 3 ?
The extreme f(x)=x^3+3x^2-24x,0<= x<= 3 is Maximum(0,0),Minimum(2,-28),Maximum(3,-18)