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# differentiating a piecewise function watch

1. (Original post by Zacken)
No, you're quite correct.

No. You say it is non-existent.
The graph of f(n) has a discnotinuity but for f'(n) it is just the line of 2n? How do you show 2n for n<1 at the point am I supposed to draw a hollow circle :I
2. (Original post by Mihael_Keehl)
The graph of f(n) has a discnotinuity but for f'(n) it is just the line of 2n? How do you show 2n for n<1 at the point am I supposed to draw a hollow circle :I
Are you sketching f(n) or f'(n)?
3. If the derivative of the function exists at 1 then must exist.

Now such a limit exists if and only if both it's left derivative and right derivative exists and are equal.

That is .

Now working with your function the left hand limit is
.

Now for (since we are approaching from the left). So try and evaluate the left hand limit and see what you get.

Remember that if the function is differentiable at 1 then we must have this left hand limit actually exists.
4. (Original post by poorform)
If the derivative of the function exists at 1 then must exist.

Now such a limit exists if and only if both it's left derivative and right derivative exists and are equal.

That is .

Now working with your function the left hand limit is
.

Now for (since we are approaching from the left). So try and evaluate the left hand limit and see what you get.

Remember that if the function is differentiable at 1 then we must have this left hand limit actually exists.
So for 1+h > 1 for h>0.

Does the right hand limit take the same form of the left hand one?

thnx btw

I suppose my reasoning is weaker but there is a discontinuity in the dx part of f(n or x) so we cannot take a gradient,

I would say that for the right hand limit it does the same thing right? The graph of f'(x) has a constant gradeint of the form 2x
5. The left side and ride side of the function derivative are not equal if you follow?

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