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October 25th, 2013, 01:08 PM   #1
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When do you replace x when finding f'(a) using the def?

The original question is to prove that

has no derivative at x=0.

The answer is apparent from the graph, but we need to prove it. The problem really boils down to taking two one-sided limits of the definition of derivative. The problem for me is that the absolute value of x+h is throwing me off.



I haven't gotten rid of the h in the denominator. I can't distribute the -2 in front of the first absolute value and I can't assume that x is 0 yet (because you're supposed to apply the limit, and then substitute for x, right?), so I can't substitute for it or figure out a comparable expression for |x+h|. Nor can I split it into three separate limits by splitting the fraction, because that would leave h on the bottom of two of them.

Note that I can get the same answer as my calculator if I substitute x=0, then finish the limit. Is my understanding that I can't replace x with 0 yet correct? If so, what's the next step?
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October 25th, 2013, 01:14 PM   #2
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Re: When do you replace x when finding f'(a) using the def?

Since you're looking for the derivative at x = 0, you can replace x by 0 right at the beginning.
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October 25th, 2013, 01:25 PM   #3
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Re: When do you replace x when finding f'(a) using the def?

Quote:
Originally Posted by Pero
Since you're looking for the derivative at x = 0, you can replace x by 0 right at the beginning.
OK. Well, as I said that matches the answer my calc gives (more accurately, doing the two derivatives that way matches the numerical derivatives that the calculator returns just to the left and right of x=0).

I suppose that if x was other than zero, the same applies.

And if we wanted to do this derivative in general, with unknown x, either I'm stuck at the above limit or we have to do a cases for x>0 and x<0. Yes?
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October 25th, 2013, 01:28 PM   #4
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Re: When do you replace x when finding f'(a) using the def?

When x is not 0, then for small enough h, x+h is positive (when x is positive) or negative (when x is negative), so you can resolve the modulus function.
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