What is the extreme f(x)=2x^3+3x^2+12 ?

The extreme f(x)=2x^3+3x^2+12 is Maximum(-1,13),Minimum(0,12)

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extreme f(x)=2x^3+3x^2+12

extreme points

f (x) = 2 x^{3} + 3 x^{2} + 12

Maximum (- 1, 13), Minimum (0, 12)

Solution steps

f^{'} (x) = 6 x^{2} + 6 x

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

2 x^{3} + 3 x^{2} + 12 : 13

Maximum (- 1, 13)

Plug

x = 0

into

2 x^{3} + 3 x^{2} + 12 : 12

Minimum (0, 12)

Maximum (- 1, 13), Minimum (0, 12)

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

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

f (x) = 2 x - y - 6

f (x) = 2 x^{2} + 3 y^{2} - 4 x + 12 y

f (x) = \frac{x ^{3}}{3} + 2 x^{2} - 4, - 2 \leq x \leq 2

What is the extreme f(x)=2x^3+3x^2+12 ?
The extreme f(x)=2x^3+3x^2+12 is Maximum(-1,13),Minimum(0,12)