What is the extreme f(x)=x^3-3x-5 ?

The extreme f(x)=x^3-3x-5 is Maximum(-1,-3),Minimum(1,-7)

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extreme f(x)=x^3-3x-5

extreme points

f (x) = x^{3} - 3 x - 5

Maximum (- 1, - 3), Minimum (1, - 7)

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 - 5 : - 3

Maximum (- 1, - 3)

Plug

x = 1

into

x^{3} - 3 x - 5 : - 7

Minimum (1, - 7)

Maximum (- 1, - 3), Minimum (1, - 7)

f (x) = 3 x^{2} + 6 x + 1

f (x) = 4 - 3 x

f (x, y) = x y^{2} - 27 x y - 27 y^{2}

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

f (x) = ln (x^{2} + 7 x + 14)

What is the extreme f(x)=x^3-3x-5 ?
The extreme f(x)=x^3-3x-5 is Maximum(-1,-3),Minimum(1,-7)