What is the extreme f(x)=x^3-12x+8 ?

The extreme f(x)=x^3-12x+8 is Maximum(-2,24),Minimum(2,-8)

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extreme f(x)=x^3-12x+8

extreme points

f (x) = x^{3} - 12 x + 8

Maximum (- 2, 24), Minimum (2, - 8)

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 + 8 : 24

Maximum (- 2, 24)

Plug

x = 2

into

x^{3} - 12 x + 8 : - 8

Minimum (2, - 8)

Maximum (- 2, 24), Minimum (2, - 8)

(arctan (y)) (tan (y))

cos (x)

f (x) = x^{2} - 8 x - 240

f (x) = - x - 2

f (x) = \frac{5 x ^{2} + 3}{x + 2}

What is the extreme f(x)=x^3-12x+8 ?
The extreme f(x)=x^3-12x+8 is Maximum(-2,24),Minimum(2,-8)