Find all values of a, B so that the function f(x)=1 if x>3, f(x)=ax-10 if x<3, and f(x)=B otherwise is continuous everywhere. I think a is continuous everywhere since ax-10 is a polynomial, and b is continuous at x=3.

Answer 1

You have the right idea. But a must be chosen so that

lim (ax-10) = 1 as x->3+

f(x) is defined every where except at x=3, so define it to be 1 there, since the limit from both sides is 1. That is,

f(3) = B

a = 11/3
B = 1