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Homework Help: New Calc student w/ a derivative question

  1. Mar 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Hello all, thank you for the help in advance. It's a two-sided derivative problem, for lack of a better term, and I appreciate all hints or help. If we have a function y so that
    y=bx for all x<0, and
    y= x^2-13x for all x> or = 0,

    for what value of b is y differentiable at all points?

    2. Relevant equations

    I would assume that d/dx (ax) = a (dy/dx), and the derivative of the part of the function greater than or equal to zero is pretty straightforward, 2x-13.

    3. The attempt at a solution

    But here, I am turning this into a pig's ear. My initial guess would be that a=x-13, since that is a factor of the second half of the equation. The derivative of that is 1, but that simply can't be right, it just doesn't pass the sniff test.

    Thank you again for your help.

  2. jcsd
  3. Mar 22, 2015 #2
    All polynomials are differentiable everywhere, so clearly you don't have to worry about any spot except the origin.

    Now go back. What's the definition of the derivative?
  4. Mar 22, 2015 #3
    Dangit, I figured it was easier than I was making it

    f(prime)(x) of x^2-13x = 2x - 13

    If x=0, then f(prime)(0) = -13

    I assume this is spot on?

  5. Mar 22, 2015 #4
    That's the derivative at 0. Now, recall that the derivative is really a limit. What must be true about the values to the left and right for the a limit to exist?
  6. Mar 22, 2015 #5
    The left and right limits must be equal to each other, so the limit is -13.
  7. Mar 22, 2015 #6


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    What you are "assuming" makes no sense. d/dx(ax)= a, there is no "y".

    Where did you get "a= x- 13"? For that matter where did you get "a"? The problem asked about a value of "b". And that may be a constant, not a function of x. You appear to be arguing that, since y= bx for x< 0, the derivative for x< 0 is the constant, b. For x> 0, y= x^2- 13x so the derivative for x> 0 is 2x- 13. Now, the derivative of a function is NOT necessarily continuous but does satisfy the "intermediate value property" so that, if y is differentiable at 0, those two "one-sided" derivatives must be the same. Do you have that theorem?

    What I would do is more fundamental than that. The definition of the derivative of function f, at x= a, is [tex]\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}[/tex]. In order that f be differentiable at x= a, that limit must exist which means two one sided limits must exist and be equal.

    Here, the limit "from the left" is [tex]\lim_{h\to 0}\frac{bh- 0}{h}= \lim_{h\to 0} b= b[/tex]. The "limit from the right" is [tex]\lim_{h\to 0}\frac{h^2- 13h- 0}{h}[/tex]. Find that limit and set it equal to b.

  8. Mar 22, 2015 #7
    Correct. So you know for sure the right hand limit is -13. You now have to make sure the left hand limit is -13. What's the derivative of the function approaching from the left side?
  9. Mar 26, 2015 #8
    I made a mistake in writing the equation -- sorry -- but I finally figured out the problem. Thank you for your help.
  10. Mar 27, 2015 #9


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    It is not generally true that the derivative of a function must be continuous but the derivative does satisfy the "intermediate value property" even when it is not continuous- that is, at some point between x= a and x= b, f'(x) must take on all values between f'(a) and f'(b). A consequence of that is that, even if f'(x) is not continuous at x= a we still must have [itex]\lim_{x\to a^+} f'(x)= \lim_{x\to a^-} f'(x)[/itex].

    Therefore, in this problem it is sufficient to require that [itex]\lim_{x\to 0^-} b= -13= \lim_{x\to 0^+} 2x- 13[/itex].
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