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September 9th, 2017, 06:31 PM   #1
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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?
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September 9th, 2017, 06:33 PM   #2
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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.
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September 10th, 2017, 04:12 AM   #3
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Quote:
Originally Posted by Leonardox View Post
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"!
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September 10th, 2017, 10:11 AM   #4
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Quote:
Originally Posted by Leonardox View Post
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.
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September 10th, 2017, 12:14 PM   #5
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$\displaystyle x^3-x-5$ has a real root because of its odd degree.
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