What is the extreme f(x)=-x^3+12x,(-6,5) ?

The extreme f(x)=-x^3+12x,(-6,5) is Minimum(-2,-16),Maximum(2,16)

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extreme f(x)=-x^3+12x,(-6,5)

extreme points

f (x) = - x^{3} + 12 x, (- 6, 5)

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

Solution steps

Find the critical points:

x = - 2, x = 2

Domain of

- x^{3} + 12 x, (- 6, 5) : - 6 < x < 5

Find intervals:

Decreasing

: - 6 < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < 5

Plug the extreme point

x = - 2

into

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

Minimum (- 2, - 16)

Plug the extreme point

x = 2

into

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

Maximum (2, 16)

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

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

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

f (x) = - \frac{5}{x - 6}

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

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

What is the extreme f(x)=-x^3+12x,(-6,5) ?
The extreme f(x)=-x^3+12x,(-6,5) is Minimum(-2,-16),Maximum(2,16)