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

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

Maximum (- 2, 3)

Plug

x = 0

into

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

Minimum (0, - 1)

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

f (x) = y + 4 x - 5

f (x, y) = \frac{x ^{2} - y ^{2}}{9 - 2 x ^{2} - y ^{2}}

f (x, y) = 9 x - 5 y + 20

f (x) = sin (x + \frac{π}{4})

8 x^{2} + 3 ln (x + 1)

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