suppose f is a continues function on point x_0
prove that g(x)=(x-x_0)*f(x) differentiable on x_0??
calculate g'(x_0)
i tried to think like this:
if f(x) is continues on x_0 then lim f(x) as x->x_0 equals f(x_0)
mvt says f'(c)=[f(a)-f(b)]
cauchys mvt says f'(c)/g'(c)=[f(a)-f(b)]/[g(a)-g(b)]
??