# Thread: MVT differentiation proof question

1. ## MVT differentiation proof question

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. Originally Posted by 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)]

??

Note, $\displaystyle \frac{g(x)-g(x_0)}{x-x_0} = \frac{(x-x_0)f(x)}{x-x_0} = f(x)$.
Also, $\displaystyle \lim_{x\to x_0}f(x)$ exists because $\displaystyle f$ is continous (in neighborhood) at $\displaystyle x_0$.

3. but they dont say that g(x) is differentiable on x_0
from where you came to this conclusion
$\displaystyle \frac{g(x)-g(x_0)}{x-x_0} = \frac{(x-x_0)f(x)}{x-x_0} = f(x)$

??

and regarding the limit you mentioned
$\displaystyle \lim_{x\to x_0}f(x)$
that means
lim f(x)=[f(x+x_0) -f(x_0)]/[x-x_0]
as x->x_0

that helps??

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

how to prove that the limits from positive side and from negative side are equal??

5. 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.

6. Originally Posted by transgalactic
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.
That is exactly what I said in post #2.
Once you cancel by $\displaystyle x-x_0$ you are left with $\displaystyle f(x)$.
Now you know that $\displaystyle \lim_{x\to x_0}f(x) = f(x_0)$ by continuity.