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

The extreme f(x)=x^3-12x+6 is Maximum(-2,22),Minimum(2,-10)

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

extreme points

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

Maximum (- 2, 22), Minimum (2, - 10)

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 + 6 : 22

Maximum (- 2, 22)

Plug

x = 2

into

x^{3} - 12 x + 6 : - 10

Minimum (2, - 10)

Maximum (- 2, 22), Minimum (2, - 10)

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

f (x) = x + \frac{2}{3}

(2.9) (7.4)

y = x - 4

\frac{( 105 )}{1 + 2 e ^{- 0.5 x}}

What is the extreme f(x)=x^3-12x+6 ?
The extreme f(x)=x^3-12x+6 is Maximum(-2,22),Minimum(2,-10)