What is the extreme f(x)=x^3-3x^2-9x+1 ?

The extreme f(x)=x^3-3x^2-9x+1 is Maximum(-1,6),Minimum(3,-26)

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extreme f(x)=x^3-3x^2-9x+1

extreme points

f (x) = x^{3} - 3 x^{2} - 9 x + 1

Maximum (- 1, 6), Minimum (3, - 26)

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 + 1 : 6

Maximum (- 1, 6)

Plug

x = 3

into

x^{3} - 3 x^{2} - 9 x + 1 : - 26

Minimum (3, - 26)

Maximum (- 1, 6), Minimum (3, - 26)

f (x) = \frac{15 x}{3 x ^{2} + 1}

f (x) = 160 x + (\frac{7200}{x})

f (x) = \frac{( 5 x + 1 )}{x - 2}

f

f (x) = ln (2 - 3 x)

What is the extreme f(x)=x^3-3x^2-9x+1 ?
The extreme f(x)=x^3-3x^2-9x+1 is Maximum(-1,6),Minimum(3,-26)