How many real roots does 2^{x} = 1 + x^{2} have?

**Solution**

Answer: 3. x = 0, x = 1 and a value just over 4.

Clearly there are no roots for negative x, since for such x, 2^{x} < 1, whereas 1 + x^{2} > 1. There are certainly roots at x = 0 and 1. Also 2^{4} < 4^{2} + 1, whereas 2^{5} > 5^{2} + 1, so there is a root between 4 and 5. We have to show that there are no other roots. Put f(x) = 2^{x} - x^{2} - 1. Then f ''(x) = (ln 2)^{2} 2^{x} - 2. This is strictly increasing with a single zero. f '(0) > 0, so f '(x) starts positive, decreases through zero to a minimum, then increases through zero. So it has just two zeros. Hence f(x) has at most three zeros, which we have already found.

© John Scholes

jscholes@kalva.demon.co.uk

22 Aug 2001