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For which a and b is the piececewise function continues everywhere

  1. Nov 6, 2011 #1
    The problem statement, all variables and given/known data:

    Sorry this is a more accurate translation:
    For which a and b does the function have a derivative everywhere

    [tex]

    f(x) = \begin{cases}
    ax+b, & x≤1\\
    x^2, & x>1
    \end{cases}[/tex]

    I'd be grateful if someone explains to me the algorithm for solving these type of problems.
    Thank you in advance.
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2
    Both ax + b and x^2 are continues functions so all you have to do is set a and b so that ax+b = x^2 when x goes to 1.
     
  4. Nov 6, 2011 #3

    Deveno

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    Science Advisor

    first, we want f to be continuous (if f is not continuous, we'll have a hard time taking the derivative, unless it's a removeable discontinuity).

    the problem obviously lies at x = 1, we should like the two definitions to agree there.

    but that gives more than one possible choice for a and b, so we should also like the derivatives of ax+b and x2 to agree at 1.

    so calculate d/dx(ax+b) and d/dx(x2) at x = 1, and take it from there...

    (in each case, the reasoning behind making the two forms of f and f' agree at x = 1, is so that the left-hand limits and the right-hand limits both agree, ensuring the continuity of f in the first case, and the differentiablity of f in the second case).
     
  5. Nov 6, 2011 #4
    So basically the limits of both parts must be the same when x is approching the point and both parts must have the same derivative at that point in order for the function to have a derivative everywhere?

    Hence lim(ax+b) as x->1, x<1 must be equal to lim(x^2) as x->1, x>1
    and (ax+b)' = (x^2)'

    I got that a = 2 and b = -1
     
    Last edited: Nov 6, 2011
  6. Nov 6, 2011 #5

    Deveno

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    Science Advisor

    well, not exactly, but it's pretty clear f(x) = ax+b is differentiable anywhere to the left of x = 1, and f(x) = x2 is differentiable anywhere to the right of x = 1.

    so the only place where the derivative might fail to exist, is at x = 1, which we can fix by choosing a and b wisely.
     
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