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'©=[f(a)-f(b)]

cauchys mvt says f'©/g'©=[f(a)-f(b)]/[g(a)-g(b)]

 

??

If you haven't learned the chain rule yet, applying the definition of the derivative should yield some good results.