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

The extreme f(x)=x^3-12x,-3<= x<= 3 is Minimum(-3,9),Maximum(-2,16),Minimum(2,-16),Maximum(3,-9)

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

extreme points

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

Minimum (- 3, 9), Maximum (- 2, 16), Minimum (2, - 16), Maximum (3, - 9)

Solution steps

Find the critical points:

x = - 3, x = - 2, x = 2, x = 3

Domain of

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

Find intervals:

Increasing

: - 3 < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = - 3

into

x^{3} - 12 x \Rightarrow y = 9

Minimum (- 3, 9)

Plug the extreme point

x = - 2

into

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

Maximum (- 2, 16)

Plug the extreme point

x = 2

into

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

Minimum (2, - 16)

Plug the extreme point

x = 3

into

x^{3} - 12 x \Rightarrow y = - 9

Maximum (3, - 9)

Minimum (- 3, 9), Maximum (- 2, 16), Minimum (2, - 16), Maximum (3, - 9)

a \cdot e^{2 x} - e^{3 x}

f (x) = 25 x^{2} e^{2 x} - 16

f (x) = e^{- 2 x^{2}}

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

f (x, y) = x^{2} + y^{2} + 5

What is the extreme f(x)=x^3-12x,-3<= x<= 3 ?
The extreme f(x)=x^3-12x,-3<= x<= 3 is Minimum(-3,9),Maximum(-2,16),Minimum(2,-16),Maximum(3,-9)