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Math Help - Finding out if a function can be derived at certain point

  1. #1
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    Finding out if a function can be derived at certain point

    Hello,

    One question that the professor of my calculus course likes to include in his exams is, if a certain function f(x) can be derived at a certain point (e.g. x=0).

    In order to find this out, do I always have to find the limit \lim\limits_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h} or can I also just find the derivative and then look at the limit \lim\limits_{x\rightarrow x_0}f'(x)?
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  2. #2
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    Quote Originally Posted by thomasdotnet View Post
    Hello,

    One question that the professor of my calculus course likes to include in his exams is, if a certain function f(x) can be derived at a certain point (e.g. x=0).

    In order to find this out, do I always have to find the limit \lim\limits_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h} or can I also just find the derivative and then look at the limit \lim\limits_{x\rightarrow x_0}f'(x)?
    The latter would seem OK. But you're best advised to go ask your professor - he's the one who will be marking your exam, not me.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    You don't always have to use the limit

    consider f(x) = x^(1/3)

    then f ' (x) =1/3 x^(-2/3) = 1/[3x^(2/3)] so f is not differentiable at 0

    Often you simply compute f ' and see where it does not exist.

    In other cases like f(x) = |x|

    then at 0 then f ' = 1 if x>0 and f ' = -1 x< 0

    here you use lim f ' at 0 to see it is not differentiable at 0

    hope this helps

    By the way we talk about whether or not a function can be differentiated
    at apoint not derived.
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  4. #4
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    Quote Originally Posted by Calculus26 View Post
    You don't always have to use the limit

    consider f(x) = x^(1/3)

    then f ' (x) =1/3 x^(-2/3) = 1/[3x^(2/3)] so f is not differentiable at 0

    Often you simply compute f ' and see where it does not exist.

    In other cases like f(x) = |x|

    then at 0 then f ' = 1 if x>0 and f ' = -1 x< 0

    here you use lim f ' at 0 to see it is not differentiable at 0

    hope this helps

    By the way we talk about whether or not a function can be differentiated
    at apoint not derived.
    I think that last point depends upon whether you speak "British" English or "American" English!
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    MHF Contributor Calculus26's Avatar
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    What ? The British speak English ?
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    Thank you guys for your help! Just to clarify this once more, is \lim\limits_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h}=\lim\limits_{x\rightarrow x_0}f'(x)?

    Quote Originally Posted by Calculus26 View Post
    Often you simply compute f ' and see where it does not exist.
    But couldn't there be a case, where the function f' isn't defined at some point, but f is still differentiable there?

    Regarding differentiable vs. derivable: Since the Americans vote for one and the British vote for the other, I guess we'll have to let the Australians decide, which one we should use.
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  7. #7
    MHF Contributor Calculus26's Avatar
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    What the above quote is saying is that f ' (x) not only exists but is continuous which is a more stringent condition.

    In fact I was in advanced calculus before i even saw an example of a function which was differentiable at a point but the derivative was not continuous at that point.

    Again you can think of functions which are defined at a point but are not

    continuous for eg f(x) = |x|/x if x does not equal 0 and = 1 if x = 0

    f(0) = 1 but f is not continuous at 0.

    as to your statement:

    But couldn't there be a case, where the function f' isn't defined at some point, but f is still differentiable there?
    the definition of f being differentiable at a point is that f ' exists at that point
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  8. #8
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    Thank you Calculus26 for that explanation, now it's a lot clearer to me.

    Quote Originally Posted by Calculus26 View Post
    Again you can think of functions which are defined at a point but are not

    continuous for eg f(x) = |x|/x if x does not equal 0 and = 1 if x = 0

    f(0) = 1 but f is not continuous at 0.
    Isn't f(x)=\frac{|x|}{x} undefined at x=0?
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