$1=1+frac1x^3$ this equation has actually no solutions due to the fact that $frac1x^3$ have the right to never equal 0 and also is undefined as soon as x=0. There fore p is false.


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We need to display that there exists one $x$ for which $x = x^3+1$. This is identical to mirroring that $f(x)$ has actually a root, where $f(x) = x^3 - x + 1$. Now use the Intermediate value Theorem.

You are watching: Is there a number that is exactly 5 more than its cube?


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Question:

"Is over there a number the is precisely one more than its cube?"(In my specific case, this was problem 51 from section 2.4 the Single variable Calculus Concepts and Contexts |4e by James Stewart)

I was just assigned this trouble in a homework assignment because that my calculus class. I took a slightly different approach compared come
Théophile, although making use of the IVT below is most likely the solution which most professors would certainly look for (especially if you were asked this question on a quiz or exam whereby you aren"t permitted a graphing calculator).

You have the right to verify the an $x$ worth which satisfies these parameters exists by graphing $y=x$ and also $y=x^3+1$ as individual functions. Over there is one intersection between these 2 graphs, that offers us the $x$ value which satisfies the equation $x=(x^3)+1$

You deserve to then plug the $x$ value right into the equation $x=(x^3)+1$ to examine the answer.

Note this method serves much more as an aid in conceptualizing this details problem, rearranging the equation $x=(x^3)+1$ and using the IVT to settle for the source is a much more exact means to a solution. That said, i initially discovered it daunting to imagine this scenario, therefore I assumed I"d share my technique in instance anyone else below is struggling in the very same way.

Cheers!


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edited january 18 "20 at 22:00
answered jan 18 "20 in ~ 21:31
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