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Math Help - some explanations

  1. #1
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    Red face some explanations

    Hello,

    I am sorry that I cannot put a more descriptive title but the fact is that the info in usual titles is what I am looking for... sort of.

    Any who, I have a certain question type to answer and I am in fact quite lost. I will give an example but I am NOT looking for an answer. I am simply looking for a guideline.

    Code:
    Find h'(2) given that f(2) = -5, f '(2) = 2, g(2) = 6, and g'(2) = 5.   
    (a) h(x) = 2f(x) - 2g(x)
    h'(2) = ???
    Cheers and thanks!

    An
    Last edited by Anhim; October 27th 2009 at 11:59 PM. Reason: code correction
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  2. #2
    Senior Member nikhil's Avatar
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    Lightbulb easy so check it out

    f(2)=-5
    f'(2)=2
    f'(2)=limit[f(2+t)-f(2)]/t
    t->0

    g(2)=6
    g'(2)=5
    g'(2)=limit[g(2+t)-g(2)]/t
    t->0

    h(x)=2f(X)-2g(x)
    h'(x)=limit[h(x+t)-h(x)]/t
    t->0
    but
    h(2+t)=2f(2+t)-2g(2+t)
    and h(2)=2f(2)-2g(2)
    therfor
    h'(x)=limit[2f(2+t)-2g(x+t)-(2f(x)-2g(x))]/t
    t->0
    or
    h'(2)= 2xlimit[f(2+t)-f(2)]/t-2xlimit[g(2+t)-g(2)]/t
    t->0 t->0
    putting the values we get
    h'(2)= 2[2-5]
    =-6
    note : due to functional equation we don't need the value of f(2) and g(2)
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  3. #3
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    Quote Originally Posted by Anhim View Post
    Hello,

    I am sorry that I cannot put a more descriptive title but the fact is that the info in usual titles is what I am looking for... sort of.

    Any who, I have a certain question type to answer and I am in fact quite lost. I will give an example but I am NOT looking for an answer. I am simply looking for a guideline.

    Code:
    Find h'(2) given that f(2) = -5, f '(2) = 2, g(2) = 6, and g'(2) = 5.   
    (a) h(x) = 2f(x) - 2g(x)
    h'(2) = ???
    Cheers and thanks!

    An
    From the linear property of the derivative (which you ought to know):

    h'(x) = 2 f'(x) - 2 g'(x) therefore h'(2) = 2 f'(2) - 2 g'(2). Now substitute the given values.
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  4. #4
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    oh my dear... How did I not see that before?!

    I knew it would be something easy as such but until you know for sure...

    Well thanks!

    An
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