What is the extreme f(x)=x^3-3x^2-1 ?

The extreme f(x)=x^3-3x^2-1 is Maximum(0,-1),Minimum(2,-5)

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extreme f(x)=x^3-3x^2-1

extreme points

f (x) = x^{3} - 3 x^{2} - 1

Maximum (0, - 1), Minimum (2, - 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{3} - 3 x^{2} - 1 : - 1

Maximum (0, - 1)

Plug

x = 2

into

x^{3} - 3 x^{2} - 1 : - 5

Minimum (2, - 5)

Maximum (0, - 1), Minimum (2, - 5)

f (x) = \frac{1}{x ^{2} - x - 6}

f (x) = x^{3} - \frac{3}{2} x^{2} - 36 x - 5

8 x y - 2 x^{4} - 2 y^{4}

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 7

f (x) = \frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + 2

What is the extreme f(x)=x^3-3x^2-1 ?
The extreme f(x)=x^3-3x^2-1 is Maximum(0,-1),Minimum(2,-5)