Mvt differentiation proof question

[transgalactic]

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)]

??

 

[quasar987]

Obviously those mvt formula you wrote don't make sense a priori since f is only assumed continuous and not differentiable.

Try doing something with the definition of the derivative of g instead.

 

[transgalactic]

the definition of a derivative is
lim f(x)=[f(x+x_0) -f(x_0)]/[x-x_0]
as x->x_0

what should i do with it?

 

[quasar987]

If you write that limit out for g, what do you get?

 

[transgalactic]

by definition if x->x_0
then the limit formula is
[g(x)-g(x_0)]/[x-x_0]

i only know this
but i don't know
what to do next?
how to combine things into a logical thing
that proves what it asks

 

[quasar987]

Well, yes, this is the formula for the derivative of g at x_0, but you also know that

g(x)=(x-x_0)*f(x).

So what do you get when you substitute this expression into the definition of g'(x_0) and take the limit?

 

[transgalactic]

i did the definition of one sided derivative
$g_ - '(x) = \mathop {\lim }\limits_{x \to x_0 - } \frac{{(x - x_0 )f'(x) - (x_0 - x_0 )f'(x)}}{{x - x_0 }} = \frac{{(x - x_0 )f(x)}}{{x - x_0 }} \\$
$g_ + '(x) = \mathop {\lim }\limits_{x \to x_0 + } \frac{{(x - x_0 )f'(x) - (x_0 - x_0 )f'(x)}}{{x - x_0 }} = \frac{{(x - x_0 )f(x)}}{{x - x_0 }}$

can i cut the numenaor and the denominator by x-x_0
and then each side will equal f(x)
so the limit is f(x_0) which is known that its continues so it equals from both side
so the limits equal

is this a correct proof.