differentiating a piecewise function Watch

Mihael_Keehl
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#21
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#21
(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
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Zacken
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(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)?
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poorform
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If the derivative of the function exists at 1 then \displaystyle f'(1)= \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} 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 \displaystyle \lim_{h\to 0+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

Now working with your function the left hand limit is
\displaystyle \lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

Now \displaystyle 1+h&lt;1 for \displaystyle h&lt;0 (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.
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Mihael_Keehl
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(Original post by poorform)
If the derivative of the function exists at 1 then \displaystyle f'(1)= \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} 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 \displaystyle \lim_{h\to 0+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

Now working with your function the left hand limit is
\displaystyle \lim_{h\to 0-}\frac{f(1+h)-f(1)}{h}.

Now \displaystyle 1+h&lt;1 for \displaystyle h&lt;0 (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
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L'Evil Wolf
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The left side and ride side of the function derivative are not equal if you follow?
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