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Differntiable at point problem

  1. Aug 23, 2009 #1
    f(x) = x/(3x + 1), prove f(x) is differentiable at point 2.

    Ok so I've had several attempts at this....

    Using Q(h) = (f(h) - f(2))/h

    I eventually end up with (h^2 -2h)/(7(3h + 1))

    Obviously the above is rubbish because it I differentiate f(x) using the normal rules then

    dy/dx = 1/(3x + 1)^2

    What am I missing here?

    Also I've tried using the difference quotient to prove it is differentiable but same result - just rubbish.

    f(c + h) - f(c)/ h

    The above also doesn't work out either! Please hep!!
  2. jcsd
  3. Aug 23, 2009 #2
    So you have the function [itex]f(x)=\frac{x}{3x+1}[/itex] and you know that a function is differentiable at [itex]a[/itex] if its derivative exists at [itex]a[/itex]. You also know that

    [tex]\left.\frac{df}{dx}\right|_{a}\equiv \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}=\frac{\frac{a+h}{3(a+h)+1}-\frac{a}{3a+1}}{h}[/tex]

    If you simplify this does it match what you expected by using the rules you know? When you take the limit does that prove the limit exists for [itex]a=2[/itex]?
  4. Aug 23, 2009 #3


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    It would be better to use Q(h)= (f(2+ h)- f(2))/h!

    Yes, because you used the wrong formula for the difference quotient.

  5. Aug 23, 2009 #4
    Jefferydk if I simplify it I end up with nothing like? Whats going on.....??????
  6. Aug 23, 2009 #5
    Err after a cup of tea and a break the penny drops......

    Thanks for your help lads! No doubt I'll call on you again!!

  7. Aug 23, 2009 #6


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    A cup of tea works wonders!

    (I used to do my best work after a couple glasses of rum- but then I could never find the papers where I had written it all down!)
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