Let h be continuous on [a, b], g continuous on [a, b] and differentable on (a, b). If g(a) = 0 and for a d not equal to zero abs[g(x)h(x) + dg'(x)] less than or equal to abs[g(x)] on [a, b], prove that g(x) =0 on [a, b]. Could someone just show me how to do this? This is close to what I posted before, but I can't figure it out and it's driving me crazy. Like the one before, I tried to use the inequality, but it doesn't make any sense. How would I do this? I'll love you to death if you can offer me some help.
Is there anyway to make it abs(d)*abs(g'(x)) <= abs(g(x)) or something like that? Obviously, I could take it from there.
Is there anyway to make it abs(d)*abs(g'(x)) <= abs(g(x)) or something like that? Obviously, I could take it from there.