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

The extreme f(x)=x^3-3x-4 is Maximum(-1,-2),Minimum(1,-6)

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

extreme points

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

Maximum (- 1, - 2), Minimum (1, - 6)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

x^{3} - 3 x - 4 : - 2

Maximum (- 1, - 2)

Plug

x = 1

into

x^{3} - 3 x - 4 : - 6

Minimum (1, - 6)

Maximum (- 1, - 2), Minimum (1, - 6)

- 5 x^{6} ln (x)

f (x) = 4 x^{3} - 6 x^{2} y + 12 x^{2} - 12 y^{2}

f (x) = 7 + 54 x - 2 x^{3}

f (x) = (2 + x)^{2} (1 - x)^{2}

y = 6 x - ln (6 x)

What is the extreme f(x)=x^3-3x-4 ?
The extreme f(x)=x^3-3x-4 is Maximum(-1,-2),Minimum(1,-6)