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My question revolves around finding a function based on its derivative of the type below :


Problem : The limit below represents the derivative of some real-valued function $f$ at some real-number $a$. State such an $f$ and $a$ in this case.

$$\lim_{h \ \to \ 0} \frac{\sqrt{9+h}-3}{h}$$


Now this type of problem is slightly unique. In general we can always find a function based on it's derivative by taking the indefinite integral of the derivative, however in this case we don't have the derivative in a general form, we only have the value for the derivative function at some point $a$, and there are a large number of $f$'s which can produce the value for the derivative at that point. Am I correct in saying this?

This problem above is easily solvable, anyone can see already that the function is obviously going to be $f(x) = \sqrt{x}$, but the way I seem to solve it is a very heuristic method, which bothers me greatly (i.e. similar to guessing a function and working from there). I'm trying to find a methodical way of solving problems of this type, as the way I solve it (shown below) will definitely break down for harder examples.


This is my solution :

By the definition of a derivative :

$$f'(x) = \lim_{h \to\ 0} \frac{f(x+h)-f(x)}{h}$$

In the above case we can see that $\ f(a+h) = \sqrt{9+h}\ $ and $\ f(a)=3 = \sqrt{9}$

This implies that $a = 9$ and $f(x) = \sqrt{x}$ or written more formally ($f : x \to \sqrt{x}, \ \forall x\in\mathbb{R^{+}}$)


As you can see that is a very hap-hazard solution, and I would like to find a better way to solve problems of these types, but it eludes me.

Is there a more methodical approach (or formal approach) to solving problems of this type, where we are given a limit representing the derivative of some real-valued function $f$ at some real-number $a$, and asked to find $f$ and $a$?

If you have any other suggestions for solving these types of problems, please comment below.

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  • $\begingroup$ I think your answer is correct and formal, taking into account that you seem to be using the definition $$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h$$ You don't seem to be needing suggestions, just lot of practice to enhance your skills of recognizing a function when given such an expression. $\endgroup$ – DonAntonio Apr 23 '16 at 15:20
  • $\begingroup$ @Joanpemo, are you saying then that there is no true methodical approach to solving problems of this type? Rather the only way to solve problems of this type is by recognition (heuristically) of a function that matches what is given in the problem. If so, then solving problems of this type would be much like evaluating integrals of functions, where there is no full-proof method. $\endgroup$ – Perturbative Apr 23 '16 at 15:31
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    $\begingroup$ Pretty much that's my opinion. Just as with integrals, differential equations and many other things in mathematics, we better do lots of exercises as there is no pre-established method for all cases, and sometimes not even for most cases. We must get as much experience as we can to recognize patterns that maybe can be worked out by some method(s) we know. $\endgroup$ – DonAntonio Apr 23 '16 at 15:34
  • $\begingroup$ To add to what Joanpemo is saying, there are a lot of standard tricks that one does to solve analysis problems. One big standard trick is to "convert the problem to a standard calculus problem". That's exactly what you're learning to do here. This trick is used all over the place. Hard ODEs/PDEs? How do you turn it into solving a bunch of integrals? Also, I don't know what led you to think there is no full-proof method for solving integrals, since there is an algorithm for it (if it can be solved in terms of elementary functions) that only requires substitution and integration by parts. $\endgroup$ – Chris Rackauckas Apr 23 '16 at 15:47
  • $\begingroup$ @ChrisRackauckas, Are you referring to the Risch algorithm? $\endgroup$ – Perturbative Apr 23 '16 at 16:18
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You are correct in saying there is a large number of maps $f$ and points $a$ satisfying this. Take any real number $a$, and any map $f$ differentiable at $a$, then this equals $g'(a)$ where $\forall x \in \mathbb{R}, g(x) = f(x) +(\frac{1}{6} - f'(a))x$. This exercice is just there for you to remember that some limits are better calculated when seen as derivatives at some point. One typical example is $\lim \limits_{x \to 0} \frac{\sin(x)}{x} = 1$, though depending on how you define $\sin$ in the first place, this one might be trivial.

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  • $\begingroup$ Is it also then correct to assume that there would be an infinite number of maps (functions) $f$ and real-numbers $a$ that could satisfy the derivative at that point $a$? $\endgroup$ – Perturbative Apr 23 '16 at 15:56
  • $\begingroup$ Yes, since for each real number $a$ and map $f$ differentiable at $a$, the map $g$ works fine and this is an injective process. So yes there is an infinite number of such maps and points; as many as there are real valued maps actually. $\endgroup$ – nombre Apr 23 '16 at 16:55

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