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Calculus 1 help

  1. Oct 29, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem is:

    f(x) = {ax^2 - b / (x - 2), x < 2
    1/4(x^2) - 3c, x >= 2

    If f is differentiable at x=2, what are the values of a, b, and c?
    In case it is hard to see, I have provided an image
    1414590365854.jpg

    2. Relevant equations

    I know that the limit definition of a function needs to be used. However that is as far as I thought.

    3. The attempt at a solution
    When finding the limit for the left side, I receive:

    lim x-> 0, xa(x+h)^2 - 2a(x+h)^2 --ax^3 -ax^2h +bh +2ax^2 / (x^2 - 4x + xh - 2h +4)h
    Unfortunately I was not able to factor out the h.

    When finding for the right side, I get -x/2 but then I can't find out how to find a, b, or c.
     
  2. jcsd
  3. Oct 29, 2014 #2

    SteamKing

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    If you have a calculus problem, don't post it in the Pre-calculus HW forum.
     
  4. Oct 29, 2014 #3
    Ah sorry about that. Thank you for moving my thread.
     
  5. Oct 29, 2014 #4

    RUber

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    Before using the definition of the derivative, start with continuity.
    The two sides have to agree at the junction. ##f(2^-)=f(2^+)##.
     
  6. Oct 29, 2014 #5
    In that case I would receive for left side:
    f(2-) = 4a - b / 2 - 2
    = 4a - b, x =/= 2


    and for right side:
    f(2+) = 1 - 3c

    So since the limits on both sides of 2 must be equivalent, I will get:
    1 - 3c = 4a - b

    But now how can I solve for each variable?
     
  7. Oct 29, 2014 #6

    Mark44

    Staff: Mentor

    Use parentheses! What you wrote is this:
    ##4a - \frac{b}{2} - 2##
    How do you figure that ##\frac{4a - b}{2 - 2} = 4a - b##?
     
  8. Oct 29, 2014 #7
    True sorry about that.
    I'm not sure what happens when ##\frac{4a - b}{2 - 2}## now as I don't know how to factor out (x-2) from the original ##\frac{4a - b}{x - 2}##
     
  9. Oct 29, 2014 #8

    Mark44

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    What does this tell you about continuity at x = 2?
     
  10. Oct 29, 2014 #9

    RUber

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    What would have to be true about ##\frac{ax^2-b}{x-2}## for you to be able to cancel out the x-2? Is this necessary for continuity?
     
  11. Oct 29, 2014 #10
    It's dicontinuous? I'm aware that "2" is actually not the value but 1.999..... but I don't know how to express that in a mathematical way.

    I'm not really sure actually. For to cancel out x - 2 it cannot be equal to 2 which it already is not according to the question right? But I'm not sure what you're implying with it being necessary for continuity.
     
  12. Oct 29, 2014 #11

    Mark44

    Staff: Mentor

    What you want to have happen is that there is a removable discontinuity at x = 2, a "hole." What values can you specify for a, b, and c, so that this is what happens?
     
  13. Oct 29, 2014 #12
    I don't understand. How can there be a "hole" if ##\frac {1} {4} x^2 - 3c## has a value at 2? Doesn't the question imply that it's a continuous function?
     
  14. Oct 29, 2014 #13

    Mark44

    Staff: Mentor

    I'm talking about the other part - ##\frac{ax^2 - b}{x - 2}##

    When x gets close to 2, from the left, the denominator gets close to zero. We don't want the graph of this part of the function to head off to infinity. That would imply a non-removable discontinuity. We won't actually have a hole when x = 2, because the other part of the function formula takes over for x >= 2, but we need to figure out the constants a, b, and c so that the curve is nice and smooth.
     
  15. Oct 29, 2014 #14

    RUber

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    Say, for example that you had ##\frac{x^2-9}{x-3}## this function is not defined at x=3, but everywhere else equivalent to x+3.
    If you had
    F(x) =##\frac{x^2-9}{x-3}## for ##x\neq 3## and 6 for x=3, this function would be continuous.
     
  16. Oct 29, 2014 #15
    Oh I see. So to make it factorable, let b = 4a and that way it becomes:
    ##\frac{a(x^2 - 4)}{x - 2}##
    ##a(x+2)##

    and now
    ##a(x + 2) = 1 - 3c##
    How can this be solved?
     
  17. Oct 29, 2014 #16

    RUber

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    Now you can put in x=2 and determine the rule for c in terms of a.
     
  18. Oct 29, 2014 #17

    RUber

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    Differentiating will lead to the unique answer.
     
  19. Oct 29, 2014 #18
    I end up with ##c = \frac{1 - 4a}{3}##
    What should I differentiate? I differentiated this to end up with:
    ##c = \frac{\frac{1 - 4a + h}{3} - \frac{1 - 4a}{3}}{h}##
    ##c = 1/3##
    but plugging that into the former gives me a = 0
     
  20. Oct 29, 2014 #19

    RUber

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    You need to differentiate the functions that you are trying to equate. The limit from the right and left should be the same for the now continuous function to also be differentiable at 2.
    Your reduced functions now should be ##a(x+2) ## and ## \frac 14 x^2 -3c##.
    This will help you solve for a, and you already have conditions for b and c in terms of a.
     
  21. Oct 29, 2014 #20
    But how can I solve for a if c is in the reduced function? Since by letting ##f(2^-) = (f(2^+)## will give me ##4a = 1 - 3c##, I still cannot solve for a due to the unknown c.
     
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