What is the extreme f(x)=x^3-3x^2-9x+5 ?

The extreme f(x)=x^3-3x^2-9x+5 is Maximum(-1,10),Minimum(3,-22)

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extreme f(x)=x^3-3x^2-9x+5

extreme points

f (x) = x^{3} - 3 x^{2} - 9 x + 5

Maximum (- 1, 10), Minimum (3, - 22)

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 + 5 : 10

Maximum (- 1, 10)

Plug

x = 3

into

x^{3} - 3 x^{2} - 9 x + 5 : - 22

Minimum (3, - 22)

Maximum (- 1, 10), Minimum (3, - 22)

f (x) = \frac{2 x + 7}{2 x - 9}

f (x) = \frac{5 x}{x - 3}

f (y) = 2 x + 5

(5.7) (- 3.1)

\frac{x - 4}{3 x + 7}

What is the extreme f(x)=x^3-3x^2-9x+5 ?
The extreme f(x)=x^3-3x^2-9x+5 is Maximum(-1,10),Minimum(3,-22)