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

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

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

extreme points

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

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

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} + b : b + 4

Maximum (- 2, b + 4)

Plug

x = 0

into

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

Minimum (0, b)

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

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

f (x) = 6 x^{2}

f (x, y) = x^{3} + y^{3} + 3 x y + 5

f (x) = e^{x^{2} - 8 x - 1}, - 8 \leq x \leq 8

f (x) = 2 x^{2} - ln (x)

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