What is the extreme f(x)=x^3-3x^2-9x+8 ?

The extreme f(x)=x^3-3x^2-9x+8 is Maximum(-1,13),Minimum(3,-19)

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extreme f(x)=x^3-3x^2-9x+8

extreme points

f (x) = x^{3} - 3 x^{2} - 9 x + 8

Maximum (- 1, 13), Minimum (3, - 19)

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 + 8 : 13

Maximum (- 1, 13)

Plug

x = 3

into

x^{3} - 3 x^{2} - 9 x + 8 : - 19

Minimum (3, - 19)

Maximum (- 1, 13), Minimum (3, - 19)

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

f (x) = x^{3} - 12 x, - 3 \leq x \leq 3

a \cdot e^{2 x} - e^{3 x}

f (x) = 25 x^{2} e^{2 x} - 16

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

What is the extreme f(x)=x^3-3x^2-9x+8 ?
The extreme f(x)=x^3-3x^2-9x+8 is Maximum(-1,13),Minimum(3,-19)