September 9th, 2017, 06:31 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 65 Thanks: 6  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^3x5 after here I think I need to plug in two consequent numbers?(how do I pick that two numbers?) to see x^3x5<0 and x^3x5>0 and? 
September 9th, 2017, 06:33 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,956 Thanks: 547 
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  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,242 Thanks: 885  Quote:
Quote:
How do you pick them? 0 is always an easy first choice, then "guess and check"!  
September 10th, 2017, 10:11 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,177 Thanks: 1644  
September 10th, 2017, 12:14 PM  #5 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
$\displaystyle x^3x5$ has a real root because of its odd degree.


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