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    I have to show at how many points a function is non analytical, one of the bits is  |x-\frac{1}{2}| . How do I do this again?
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    I would presume by contradiction, assuming it is differentiable and using the relevant definitions to try and prove this, and subsequently using some contradiction which hence arises to demonstrate it is not...?

    I've also requested this be moved to study help as one of the helpers there might be better able to run through any issues you have
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    (Original post by BinaryJava)
    I have to show at how many points a function is non analytical, one of the bits is  |x-\frac{1}{2}| . How do I do this again?
    Is the point where the function is not differential error clear to you? How would you try to work out the gradient/gradients there?
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    (Original post by RichE)
    Is the point where the function is not differential error clear to you? How would you try to work out the gradient/gradients there?
    I know that when x=1/2 you can't define a gradient geometrically, but how do you show it. I used the limit definition and ended up with  \frac{|h|}{h} . This would mean different limits for upper and lower, but you get this result for any x you choose, but it should only fail for x=1/2. Is this because when you sub in other x values you can remove the modulus?
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    (Original post by BinaryJava)
    I know that when x=1/2 you can't define a gradient geometrically, but how do you show it. I used the limit definition and ended up with  \frac{|h|}{h} . This would mean different limits for upper and lower, but you get this result for any x you choose, but it should only fail for x=1/2. Is this because when you sub in other x values you can remove the modulus?
    Find limit as x approaches 1/2 from the left and from the right and show that they're different. Limit exists if and only if limit from left is same as the limit from the right, so if no limit exists then the function is not differentiable at that point.
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    (Original post by BinaryJava)
    I know that when x=1/2 you can't define a gradient geometrically, but how do you show it. I used the limit definition and ended up with  \frac{|h|}{h} . This would mean different limits for upper and lower, but you get this result for any x you choose, but it should only fail for x=1/2. Is this because when you sub in other x values you can remove the modulus?
    No you don't - you would get a gradient of 1 from upper and lower when (say) x=1.
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    Where have I gone wrong
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    (Original post by BinaryJava)
    Where have I gone wrong
    The absolute value of 1/2 + h equals 1/2+h, when h is small, and whether or not h is positive or negative
 
 
 
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