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Home Q & A Board Science-Related Homework Help Mathematics
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bucvet79 bucvet79
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Posts: 29
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10 years ago
so for number 57 part of my chapter review (test tomarrow) it says use a graphing utility to graph the function and determine the open intervals on which the function is increasing, decreasing, or constant.

so the equation is f(x)=x^3-3x. What i dont understand is what the heck their talking about the open intervals and the increasing decreasing or constant
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#1
10 years ago
You won't need a graphing calculator.

f ' (x ) = 3x^2 - 3

if f ' ( x ) > 0; then it's increasing
if f ' ( x ) < 0; then it's decreasing

so 3x^2 - 3 > 0
  x^2 - 1 > 0
(x - 1)(x +1) > 0
therefore, x > 1 and x < - 1 is where it increases.


   3x^2 - 3 < 0
   x^2 - 1 < 0
  (x - 1)(x + 1) < 0
 so , it decreases when - 1 < x < 1

and is constant when x = -1 or x = 1
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_Mil96_Mil96
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#2
Posts: 64
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10 years ago
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