What is the extreme f(x)=x^3-3x^2-9x+12 ?

The extreme f(x)=x^3-3x^2-9x+12 is Maximum(-1,17),Minimum(3,-15)

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extreme f(x)=x^3-3x^2-9x+12

extreme points

f (x) = x^{3} - 3 x^{2} - 9 x + 12

Maximum (- 1, 17), Minimum (3, - 15)

Solution steps

f^{'} (x) = 3 x^{2} - 6 x - 9

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

x^{3} - 3 x^{2} - 9 x + 12 : 17

Maximum (- 1, 17)

Plug

x = 3

into

x^{3} - 3 x^{2} - 9 x + 12 : - 15

Minimum (3, - 15)

Maximum (- 1, 17), Minimum (3, - 15)

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12, (0, 1)

f (x, y) = x^{2} + y^{2} + 12 x - 16 y

f (x, y) = \frac{3}{x ^{2} + y ^{2} - 9}

6 x + 12

- 2 x^{2} - x + 3

What is the extreme f(x)=x^3-3x^2-9x+12 ?
The extreme f(x)=x^3-3x^2-9x+12 is Maximum(-1,17),Minimum(3,-15)