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

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

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

extreme points

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

Maximum (- 2, 1), Minimum (0, - 3)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

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

Maximum (- 2, 1)

Plug

x = 0

into

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

Minimum (0, - 3)

Maximum (- 2, 1), Minimum (0, - 3)

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

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

f (x) = \frac{2 x + 7}{5 x}

\frac{x}{x + 4}

y = 36 - x^{2}

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