Derivative of a FunctionDate: 10/05/2001 at 12:59:22 From: Komol Flood Subject: Derivatives The problem looks easy enough. I am to find the derivative of the function (x-6)(x+1)/(x-6) at x=6. I simplified the function to a linear function x+1 with a "hole" at x=6. Then I tried to take the derivative. Since the derivative is a limit, I wasn't too concerned about the "hole" at x=6 and I found the value to be ONE. I like this answer and the logical way I answered the question and used the definition of the derivative seemed to solidify my understanding of the concepts. Now for the bad part. My TI-89 calculator says division by zero, infinite result. My friend claims the derivative is the slope of the line tangent to the graph at that point (true). But how can there be a line tangent at that point if the point isn't defined? I cannot see the flaw in her argument, nor can I see the flaw in mine, but the two are incompatible. Any chance you could find the time to explain the flaw in one of our reasonings? I hope so, because this simple problem has really created some conceptual problems. Date: 10/05/2001 at 14:54:25 From: Doctor Douglas Subject: Re: Derivatives Hi Komol, and thanks for writing. Your friend is correct. A function must be defined at a given point in order for its derivative to exist there. To see why, let's examine the definition of the derivative as a limit: f(x) - f(a) f'(a) = lim ----------- x->a x - a As you can see, this definition requires the evaluation of the function f precisely at the the point x=a. The definition does not say something like f(x)-f(b)/(x-b) where x and b both approach a (and x not equal to b, of course). It's that pesky f(a) in the numerator of the limit above that prevents us from computing the limit at f(a) in your problem. Of course, your function has derivative f' equal to +1 everywhere except at x=6. We could DEFINE the value of the function f at a=6 to be f(a)=7, but this is adding more information to the problem than we were originally given. If we are permitted to do that, then the function does have a derivative that is equal to +1 everywhere. Now, we could also DEFINE the value of the function f at a=6 to be something else, say f(6)=22. This would fill in the "hole" with a different value, and now the function is defined at f(6), but if you attempt to compute the limit using f(6)=22 for f(a) in the fraction above, the limit (and hence the derivative) does not exist at x=a. I hope this helps. Please write back if you have any more questions about this. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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