Find all vales for a so that
f(x) { x+1...x<=a
{ x^2...x>a
is continuous
well , at a, both functions have the same value, so
x+1=x^2
a+1=a^2
a^2-a-1=0
solve for a. use the quadratic equation. I think this involves the golden ratio: http://mathworld.wolfram.com/GoldenRatio.html