What is the extreme f(x)=x^3+3x^2-4 ?

The extreme f(x)=x^3+3x^2-4 is Maximum(-2,0),Minimum(0,-4)

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extreme f(x)=x^3+3x^2-4

extreme points

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

Maximum (- 2, 0), Minimum (0, - 4)

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} - 4 : 0

Maximum (- 2, 0)

Plug

x = 0

into

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

Minimum (0, - 4)

Maximum (- 2, 0), Minimum (0, - 4)

f (x) = 0.2 x^{2} + \frac{1}{2 x}

(- 6, 1), (1, 8)

f (x) = \frac{4 x + 9}{7 x}

f (x) = 2 x^{5} + x^{7}

What is the extreme f(x)=x^3+3x^2-4 ?
The extreme f(x)=x^3+3x^2-4 is Maximum(-2,0),Minimum(0,-4)