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Homework Help: Derivative: in trouble.

  1. Jan 14, 2005 #1
    Ok, they tell me tofind a and b so that the function:
    f(x)= *Root(2-x^2) if -root(2)<=x<=1
    *x^2 + ax + b if x>1
    has derivative at 1
    I got that the condition for this graph to be continous at 1 is a+b=0
    And I moved to check out the derivative stuff:
    When x->1+, the derivative is 0
    But when 1->1-, igot stuck:
    lim (x->1-) of [f(x)-f(1)]/(x-1) = lim (x->1-) of (x^2+ax+b-1)/(x-1). So how should I move on?
  2. jcsd
  3. Jan 14, 2005 #2


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    Equate the lateral limits for the derivative and get another relation between "a" and "b".
    Then use the 2 equations found to determine "a" and "b".

  4. Jan 14, 2005 #3


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    How did u compute those derivatives??Post your work.My advice from the previous post remains valid as soon as u dig those derivatives rightly.

  5. Jan 14, 2005 #4
    We'll check when x>1:
    lim (x->1+) of [f(x)-f(1)]/(x-1) = lim (x->1+) of [x^2+ax+b-1]/(x-1)] (from the function at my first post, I got that when x>1, f(x)=x^2+ax+b)
  6. Jan 14, 2005 #5
    Come on, anybody help me out!
    Last edited: Jan 14, 2005
  7. Jan 14, 2005 #6


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    Do you really need to use the basic formula for a derivative? You should know, by the time you are doing a problem like this, that the derivative of x2+ax+b is
    2x+ b.
    That tells you that the derivative, as you approach 1 from the right, is 2+ b, not 0.

    The derivative of [itex]\sqrt{2- x^2}= (2- x^2)^{\frac{1}{2}}[/itex] is [itex]\frac{1}{2}(2-x^2)^{-\frac{1}{2}}(-2x)[/itex] which is -1 at x= 1. That is, the derivative, as you approach from the left is -1. In order that f be differentiable at x= 1, you must have 2+ b= -1.
  8. Jan 14, 2005 #7
    Yup, my class is at that lession, tis is an apllication. So, I didn't really understand you.:-) But I'll try to read it more.
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