0
$\begingroup$

Consider the following question

question

I would attempt to answer this as follows:

If the derivative of $f$ exists at $a \in (b,c)$, it would be defined as follows

$$ f'(a) = \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} $$

And since $f(a) \equiv g(a)$ for all $a \in (b,c)$, we can say that

$$ f'(a) \equiv \lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a} = g'(a) > $$

and since we are given that $g'(a)$ is defined for all $a \in (b,c)$, we know that $f'(a)$ must be defined for all such $a$.

However, the actual solution provided for this question is as below

sol

So is my attempted proof insufficient? If so, why?

$\endgroup$
  • $\begingroup$ You cannot say that $f'(x)=...$ as you don't know if it exists.To avoid this mistake start from $g'(x)$ and show that $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ exists. Except for the above mentioned you are correct. $\endgroup$ – Andreas Ch. Feb 28 '17 at 0:43
  • $\begingroup$ Many books on calculus try to over complicate things. The statement is too trivial to be proved explicitly and your thinking is correct in this matter. $\endgroup$ – Paramanand Singh Feb 28 '17 at 9:45
4
$\begingroup$

Your approach is correct. The task is just to show that existence of derivative is a local property (it is exactly saying that change outside of a neighborhood does not change the property), and by definition of derivative it suffices to show that an existence of limit is a local property.

Now it begins to be so obvious that it is difficult to say what should be proven and what can be just stated. If we want to show that existence of limit is a local property, we expand the definition of limit and use $\delta < \min(a-b, c-a)$, so $f$ and $g$ restricted to $(a-\delta, a+\delta)$ are equal.

I do not know why your source argues it with sequences. Perhaps the author likes the view of limits via sequences and he wants to practice it, or he just did not realize at the moment how trivial the fact is. But I encourage you in simplifying given proofs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.