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Piecewise function - Find derivative at 3

  1. Oct 20, 2009 #1
    In this question, we shall take steps to find the values of a and b , given that the function

    f(x)={

    x^2−4x+1 if x<=3
    ax+b if x>3

    is differentiable at 3.

    a) It is known that if a function is differentiable at a point c, then it is continuous at c. Using now the continuity of f at 3, we can establish a relationship between a and b. Find this relationship and express it in the form b=Aa+B, where A and B are constants.


    b) Assuming that x>3, one can simplify the quotient

    f(x) -f(3)
    x-(3)

    into the form Ca+D, where C and D are constants. Find these constants.

    Hint. Don't forget that you can use the result from part (a) to eliminate b from your expression.


    (c) Assuming that x<3, one can simplify the quotient

    f(x) -f(3)
    x-(3)

    into the form Ex+F, where E and F are constants. Find these constants.


    (d) Using the results of parts (a), (b) and (c), find the values of a and b.


    Answers:
    I got part:

    a)
    A=0
    B=-2

    b)
    C=1
    D=1

    c)
    E=1
    F=-1

    d)
    a=-2/3
    b=0


    However, I couldn't get correct answers on "A" from part a) and "a" and "b" from part d). Can anybody tell me what I did wrong?
     
    Last edited: Oct 21, 2009
  2. jcsd
  3. Oct 20, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    Re: Differentiation

    showing your working will help find where you went wrong
     
  4. Oct 20, 2009 #3
    Re: Differentiation

    Part A is something like this:

    ax+b
    a(3)+b = 0

    ==>b = -3a - 2

    f(x) = x^2-4x+1
    f(3) = (3)^2-4(3)+1 = -2*

    f'(x) = 2x-4
    f'(3) = 2(3)-4-2* = 0

    if x=3, then
    f’(x) = a, which must be also = 0
    thus a=0, hence b=-3a-2 = 3(0)-2 = -2

    So, what did I do wrong here?
     
  5. Oct 21, 2009 #4

    Mark44

    Staff: Mentor

    Re: Differentiation

    The next line is wrong. Why are you subtracting 2? What is the significance of the asterisk?
     
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