Dr. Gupta gave a couple of very important hints about the upcoming midterm exam for Math 1500 (and she is not one to mislead students).

 

First, she said know how to prove a function, f(x) is not differentiable at x = 6 (or whatever value they choose).  She mentioned the alternative definition of derivative formula which I mention in question 4 of my Definition of Derivative Practise Problems (page 106).  This formula states:

 

f ' (a) = the limit as x approaches a of [ f(x) - f(a) ] / [ x - a ]

 

If the limit exists, f ' (a) exists, proving that f(x) is differentiable at x = a.  If the limit does not exist, that proves f(x) is not differentiable at x = a.

 

Personally, I dislike using this formula because it makes for more challenging algebra, as I show in my question 4.  Instead, I would always revert to the classic definition of derivative formula, but replace the x with the specific value a:

 

f ' (a) = limit as h approaches 0 of [ f(a + h) - f(a) ] / h

 

However, I suspect that this question will arise as part of a continuity problem where you are given a piecewise function. In that case, there is a much easier way to deal with the problem.

 

First, one of the required theorems you have to be able to prove is "differentiability implies continuity".  This means that, if a function is not continuous at x = a, then there is no way it can be differentiable at x = a.  However, if the function is continuous at x = a, it may still not be differentiable at x = a.

 

Assuming I am right, I believe they will give you a piecewise function and first ask you to prove f(x) is continuous at x = 6 (or whatever).  You, of course, will do so the usual way, as I teach in Lesson 3 (Continuity).

 

They will then ask if f(x) is differentiable at x = 6 (or whatever).  First, if you have proven already that f(x) is not continuous at x = 6, then you would simply say, "No, f(x) is not differentiable at x = 6 because the function is not continous at x = 6."

 

However, if you have proven that f(x) is continous at x = 6, then you will say:

 

f(x) is differentiable at x = a if and only if

the limit as x approaches a- of f ' (x) =  the limit as x approaches a+ of f ' (x)

 

Which is to say, compute f ' (x) for each piece of the piecewise function and do the limit of the appropriate piece of f ' (x) as you approach a from a specific side.  If the left limit and the right limit are the same, that proves f ' (a) exists and equals that value.  If the limits do not equal, that proves f ' (a) does not exist ( f(x) is not differentiable at x = a).

 

Here is a link to an example of this type of problem:

Checking Differentiability