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Derivative of a Function

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Date: 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.
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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

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
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

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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