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

The extreme f(x)=x^3+3x^2+4 is Maximum(-2,8),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, 8), 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 : 8

Maximum (- 2, 8)

Plug

x = 0

into

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

Minimum (0, 4)

Maximum (- 2, 8), Minimum (0, 4)

f (x, y) = x^{2} + x y + y^{2} + 4 y

f (x) = \frac{x}{9 + x ^{2}}

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

f (x) = 2 x^{3} + 9 x^{2} - 24 x

What is the extreme f(x)=x^3+3x^2+4 ?
The extreme f(x)=x^3+3x^2+4 is Maximum(-2,8),Minimum(0,4)