What is the extreme f(x)=2x^3+12x^2+18x ?

The extreme f(x)=2x^3+12x^2+18x is Maximum(-3,0),Minimum(-1,-8)

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extreme f(x)=2x^3+12x^2+18x

extreme points

f (x) = 2 x^{3} + 12 x^{2} + 18 x

Maximum (- 3, 0), Minimum (- 1, - 8)

Solution steps

f^{'} (x) = 6 x^{2} + 24 x + 18

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 3

into

2 x^{3} + 12 x^{2} + 18 x : 0

Maximum (- 3, 0)

Plug

x = - 1

into

2 x^{3} + 12 x^{2} + 18 x : - 8

Minimum (- 1, - 8)

Maximum (- 3, 0), Minimum (- 1, - 8)

f (x) = \frac{4}{3} x^{3} + 7 x^{2} - 8 x - 42

f (x) = 2 x 16 - x^{2}

f (x) = 16 - x^{2} - y^{2}

f (x, y) = 25 x^{2} + 4 y^{2} + 4

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

What is the extreme f(x)=2x^3+12x^2+18x ?
The extreme f(x)=2x^3+12x^2+18x is Maximum(-3,0),Minimum(-1,-8)