What is the extreme f(x)=2x^3+6x^2-18x ?

The extreme f(x)=2x^3+6x^2-18x is Maximum(-3,54),Minimum(1,-10)

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extreme f(x)=2x^3+6x^2-18x

extreme points

f (x) = 2 x^{3} + 6 x^{2} - 18 x

Maximum (- 3, 54), Minimum (1, - 10)

Solution steps

f^{'} (x) = 6 x^{2} + 12 x - 18

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 3

into

2 x^{3} + 6 x^{2} - 18 x : 54

Maximum (- 3, 54)

Plug

x = 1

into

2 x^{3} + 6 x^{2} - 18 x : - 10

Minimum (1, - 10)

Maximum (- 3, 54), Minimum (1, - 10)

cos (x^{2}) + 5 cot (x)

f (x) = \frac{6 x ^{2}}{x ^{2} - 4}

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

x^{2} - 4 x + 3

f (x) = \frac{5}{2 x + 3}

What is the extreme f(x)=2x^3+6x^2-18x ?
The extreme f(x)=2x^3+6x^2-18x is Maximum(-3,54),Minimum(1,-10)