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 September 9th, 2017, 06:31 PM #1 Newbie   Joined: Apr 2017 From: New York Posts: 22 Thanks: 2 Intermediate Value Theorem question: Is there a number that is exactly 5 more than its cube? What I tried : x=number x+5=x^3 if we rearrange: x^3-x-5 after here I think I need to plug in two consequent numbers?(how do I pick that two numbers?) to see x^3-x-5<0 and x^3-x-5>0 and?
 September 9th, 2017, 06:33 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,527 Thanks: 364 Try 0 and a trillion. How'd I guess those? You can see that for x near 0, the -5 predominates. And for x large, the x^3 predominates. But even x = 2 works since 2^3 - 2 = 6.
September 10th, 2017, 04:12 AM   #3
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Quote:
 Originally Posted by Leonardox question: Is there a number that is exactly 5 more than its cube? What I tried : x=number x+5=x^3 if we rearrange: x^3-x-5
You've lost the "=". You should have x^3- x- 5= 0.

Quote:
 after here I think I need to plug in two consequent numbers?(how do I pick that two numbers?) to see x^3-x-5<0 and x^3-x-5>0 and?
What do you mean by "consequent" here? If you are thinking they must be successive integers, n and n+1, that is not correct. It is not hard to see that if x= 0, 0^3- 0- 5= -5< 0 and that if x=2, 2^3- 2- 5= 8- 2- 5= 1> 0.

How do you pick them? 0 is always an easy first choice, then "guess and check"!

September 10th, 2017, 10:11 AM   #4
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Quote:
 Originally Posted by Leonardox Is there a number that is exactly 5 more than its cube?
Yes. It's close to -1.904160859 as -1.904160859³ + 5 = -1.90416086 approximately.

 September 10th, 2017, 12:14 PM #5 Member   Joined: May 2017 From: Russia Posts: 33 Thanks: 4 $\displaystyle x^3-x-5$ has a real root because of its odd degree.

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