# Mvt differentiation proof question

1. Jan 20, 2009

### 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)]

??

2. Jan 20, 2009

### 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.

3. Jan 21, 2009

### 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?

4. Jan 21, 2009

### quasar987

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

5. Jan 21, 2009

### 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 dont know
what to do next?
how to combine things into a logical thing
that proves what it asks

6. Jan 21, 2009

### 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?

7. Jan 30, 2009

### 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) wich is known that its continues so it equals from both side
so the limits equal

is this a correct proof.

Last edited: Jan 30, 2009
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