What is the extreme f(x)=x^3-12x+4 ?

The extreme f(x)=x^3-12x+4 is Maximum(-2,20),Minimum(2,-12)

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extreme f(x)=x^3-12x+4

extreme points

f (x) = x^{3} - 12 x + 4

Maximum (- 2, 20), Minimum (2, - 12)

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 + 4 : 20

Maximum (- 2, 20)

Plug

x = 2

into

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

Minimum (2, - 12)

Maximum (- 2, 20), Minimum (2, - 12)

y = e^{r x}

\frac{- 2 x ^{2} + 5 x - 1}{2 x - 1}

f (x) = x^{3} - 6 x^{2} + 9 x + 2

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

f (x) = 3 x^{4} - 4 x^{3} + 1

What is the extreme f(x)=x^3-12x+4 ?
The extreme f(x)=x^3-12x+4 is Maximum(-2,20),Minimum(2,-12)