What is the extreme f(x)=x^3-12x,0<= x<= 5 ?

The extreme f(x)=x^3-12x,0<= x<= 5 is Maximum(0,0),Minimum(2,-16),Maximum(5,65)

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extreme f(x)=x^3-12x,0<= x<= 5

extreme points

f (x) = x^{3} - 12 x, 0 \leq x \leq 5

Maximum (0, 0), Minimum (2, - 16), Maximum (5, 65)

Solution steps

Find the critical points:

x = 0, x = 2, x = 5

Domain of

x^{3} - 12 x, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 5

Plug the extreme point

x = 0

into

x^{3} - 12 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 2

into

x^{3} - 12 x \Rightarrow y = - 16

Minimum (2, - 16)

Plug the extreme point

x = 5

into

x^{3} - 12 x \Rightarrow y = 65

Maximum (5, 65)

Maximum (0, 0), Minimum (2, - 16), Maximum (5, 65)

f (x) = 2 x^{3} - 6 x^{2} + 4

f (x) = 3 x - 9 x^{\frac{1}{3}}

g (x, y) = c x + y x

x^{3} + 6 x^{2} + 9 x

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

What is the extreme f(x)=x^3-12x,0<= x<= 5 ?
The extreme f(x)=x^3-12x,0<= x<= 5 is Maximum(0,0),Minimum(2,-16),Maximum(5,65)