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fallout
11-08-2002, 11:42 AM
Ok, I got stumped by a co-worker on this one today.

She described it like this.

A gym charges by the hour. \$2 for the first hour. \$1 the next hour, \$1 the hour after that.

the function would look something like

f(x) = 2 0&lt;x&lt;=1, f(x) = 3 1&lt;x&lt;=2, f(x) = 4 2&lt;x&lt;=3,....

The question was which graph represents the derivative?

The choices that made sense are f'(x) = 0 defined everywhere, and a second one f'(x) = 0 everywhere, except undefined for x=1,2,3...

So which one was it?

It is definetly the day after so I hope I have not posted this too early.

Gandalf
11-08-2002, 12:03 PM
Undefined for x=1,2,3...

Derivative exists only if it exists for small changes in both directions, and those two results are equal. At least that's what my calc class said, long ago.

fallout
11-08-2002, 12:21 PM
Undefined for x=1,2,3...

Derivative exists only if it exists for small changes in both directions, and those two results are equal. At least that's what my calc class said, long ago.

But the derivative from the left = the derivative from the right does it not?

I always thought continuity of the function was not a requirement for the derivative to be defined. My Calculus was many eons ago also.

from the earlier post:

"the function would look something like
f(x) = 2 0&lt;x&lt;=1, f(x) = 3 1&lt;x&lt;=2, f(x) = 4 2&lt;x&lt;=3,.... "

therefore f'(x) = 0 0&lt;x&lt;=1, f'(x) = 0, 1&lt;x&lt;=2, ...

therefore f'(x) = 0 always?

Gandalf
11-08-2002, 12:32 PM
My calc course definition of f'(x) = lim h goes to 0 of (f(x+h)-f(x))/h.

Since for small positive values of h, lim (f(1+h) - f(1)) / h does not exist, while for small negative values of h it does, left and right cannot be equal.

This means that to be differentiable at x, it must be continuous at x. The converse is not true.

[I think special rules would apply at endpoints. E.g., a function defined only for nonnegative x could be differentiable at 0. Here, the function is defined on [0, infinity) so at points like 1 it must be differentiable both ways.]

fallout
11-08-2002, 12:53 PM
I asled her if there was a 0 at 0 and she said she doesn't think so.

If the function she describes had no circle around 0, then there would be an error in the question would there not be?

This:

----0---0---0---&gt;
0 1 2 3...

Couldn't possibly be right could it?

I told her

-----------&gt;
0 1 2 3

Makes more sense since there was no 0 at 0.

I think I will tell her to question the question. She chose the one with no 0's.

Do you think they will say the one with o's at 1, 2, 3 is still better, and they say choose the "best" answer, even if it is obviously wrong?

Gandalf
11-08-2002, 01:08 PM
It seems OK to question it, but she'll almost certainly lose.

There seem to be two possible arguments the SOA could claim (and they don't even have to tell you why they reject a challenge), in addition to the "choose the best answer" argument:

Both relate to the fact that you came up with the formula for f(x), not them. Possibilities:

1. f is defined at 0, and f(0) = 2. Then since the domain of f does not include negative x, you only need to consider the right derivative, which is 0.

2. f is not defined at 0, just as it is not defined for x&lt;0. You cannot make a deal with the gym that involves no time. If 0 is not in the domain of f, you shouldn't need to explicitly exclude it from a graph of the derivative of f.

sassytsw
11-08-2002, 01:37 PM
I am looking at the Averbach/Mehta manual now (Chapter 2, page 2) which defines differentiability on piecewise functions. In order to be differentiable at a point x=a, the function must be continuous at x=a as well as the limit of x as it goes to a from the left equaling the limit of x as it goes to a from the right. In the exam problem, neither of these conditions are satisfied due to the jump discontinuity of the function. Therefore, the derivative does not exist at x=1, 2, etc. Choice E would be the best answer.

Carol Burnett
11-08-2002, 01:43 PM
man, this frustrates me! :( I had originally chosen the option with the 0's because of the limit from the right not being equal to the limit from the left. But I ended up changing it to the straight zero vector. Now that I'm thinking it through, i'm not for sure what my exact reason for changing was. oh well.