Question

Is there any real number exactly one less than its cube intermediate value theorem?

Answer 1

Answer:

≈ 1.32471795725...

Explanation:

If x is one less than its cube, then

x = x³ - 1,

x³ - x - 1 = 0

so f(x) = ? would be an appropriate (continuous) function to apply the Intermedate Value Theorem on some appropriate interval to see if it takes on the value 0 in that interval.

f(x) = x³ - x - 1

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:

$\sqrt[3]{\frac{1}{2} + \sqrt{\frac{1}{4} - \frac{1}{27} } } + \sqrt[3]{\frac{1}{2} - \sqrt{\frac{1}{4} - \frac{1}{27} } }$

≈ 1.32471795725...