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

The extreme f(x)=x^3-12x+12 is Maximum(-2,28),Minimum(2,-4)

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extreme f(x)=x^3-12x+12

extreme points

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

Maximum (- 2, 28), Minimum (2, - 4)

Solution steps

f^{'} (x) = 3 x^{2} - 12

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{3} - 12 x + 12 : 28

Maximum (- 2, 28)

Plug

x = 2

into

x^{3} - 12 x + 12 : - 4

Minimum (2, - 4)

Maximum (- 2, 28), Minimum (2, - 4)

y = - 3 x + 4

120 x - 0.4 x^{4} + 700

x + y = - 4

x^{2} + 2 x

f (x) = \frac{x}{6 x + 25}

What is the extreme f(x)=x^3-12x+12 ?
The extreme f(x)=x^3-12x+12 is Maximum(-2,28),Minimum(2,-4)