Is any real number exactly 1 less than its cube?

[BadAtCalc]

Is any real number exactly 1 less than its cube? Answer this question by applying the intermediate value property (you need to determin a suitable finction and a suitable interval)

I really dont understand, can someone help?

Thanks

 

[Unco]

If x is one less than its cube, then x = x^3 - 1, so f(x) = ? would be an appropriate (continuous) function to apply the IVP on some appropriate interval to see if it takes on the value 0 in that interval.

 

[Count Iblis]

Is any real number exactly 1 less than its cube? Answer this question by applying the intermediate value property (you need to determin a suitable finction and a suitable interval)

I really dont understand, can someone help?

Thanks

x^3-x-1=0

For large x the left hand side is positive, for x = 0 it is negative. The root can be calculated exactly, it is given by:

\(\displaystyle \L\sqrt[3]{\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{1}{27}}}+\sqrt[3]{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{1}{27}}}\approx 1.32471795725\cdots\)