Verified answer

Hi Joe.

Without loss of generality assume f(a) >0.

By continuity of f it follows there exists d>0 such that if |x-a|< d then |f(x) - f(a) | < |f(a)|.

This implies -|f(a)| < f(x) - f(a) < |f(a)| so in particular f(a) - |f(a)| < f(x) < |f(a)| + f(a).

Since f(a) >0 then |f(a)| = f(a) hence we obtain:

f(a) - f(a) < f(x) < 2f(a)

0 < f(x) < 2f(a)

So in particular f(x) >0 (and thus f(x) ̸=0) for all x such that |x-a|< d, i.e for all x in the interval (a - d, a+ d).

Therefore it suffices to take r = d.