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

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

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

extreme points

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

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

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} + 1 : 5

Maximum (- 2, 5)

Plug

x = 0

into

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

Minimum (0, 1)

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

3 x - 5 y = 4

- (\frac{1}{3})^{x}

x^{2} - 2 x + 3

f (x) = 2 x + 13

(0, 2765), (350, 8890)

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