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

The extreme f(x)=x^3+3x^2+2 is Maximum(-2,6),Minimum(0,2)

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

extreme points

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

Maximum (- 2, 6), Minimum (0, 2)

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} + 2 : 6

Maximum (- 2, 6)

Plug

x = 0

into

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

Minimum (0, 2)

Maximum (- 2, 6), Minimum (0, 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)

f (x) = 36 - x^{2} - y^{2}

What is the extreme f(x)=x^3+3x^2+2 ?
The extreme f(x)=x^3+3x^2+2 is Maximum(-2,6),Minimum(0,2)