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Math Help - Derivative of a function

  1. #1
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    Derivative of a function

    A function f is defined for all real numbers and has he following three properties.
    1) f(1)=5
    2) f(3)=21
    3)for all real values of a and b, f(a+b)-f(a)=kab+2b^2

    a)use a=1 and b=2 to find the value of k
    f(1+2)-f(1)=k(1)(2)+2(2)^2
    f(3)-f(1)=2k+8
    21-5=2k+8
    8=2k
    k=4
    Is that right?

    B)find f'(3)
    c) find f'(x) and f(x) for all real x
    I need help figuring out how to do the other 2 parts.
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  2. #2
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    If I am interpreting this correctly:

    Part A: Not sure, forget how to grab a 3rd point..and construct the equation from there

    Parts B and C: These are easy once you get the equation of a function (i.e. f(x) ). Once you have that, you can power rule to get the derivative and sub 3 into the derivative of the function to get your answer.

    Ibrox
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  3. #3
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    Quote Originally Posted by calc_student09 View Post
    A function f is defined for all real numbers and has he following three properties.
    1) f(1)=5
    2) f(3)=21
    3)for all real values of a and b, f(a+b)-f(a)=kab+2b^2

    a)use a=1 and b=2 to find the value of k
    f(1+2)-f(1)=k(1)(2)+2(2)^2
    f(3)-f(1)=2k+8
    21-5=2k+8
    8=2k
    k=4
    Is that right?

    Mr F says: Yes.

    B)find f'(3)
    c) find f'(x) and f(x) for all real x
    I need help figuring out how to do the other 2 parts.
    Let a = b = x: f(2x) - f(x) = 6x^2 \Rightarrow f(x) = 2x^2 + c where c is a constant.
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  4. #4
    Member OnMyWayToBeAMathProffesor's Avatar
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    I also got k=4. But how would you do f'(3)? and i am sorry 'mr fantastic', i did not understand your process. could you please explain?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Let a = b = x: f(2x) - f(x) = 6x^2 \Rightarrow f(x) = 2x^2 + c where c is a constant.
    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I also got k=4. But how would you do f'(3)? and i am sorry 'mr fantastic', i did not understand your process. could you please explain?
    Given f(2x) - f(x) = 6x^2, it's not unreasonable to assume that f(x) has the form \alpha x^2. Substitute this into f(2x) - f(x) = 6x^2 to find that \alpha = 2.

    Once you have the rule it's simple to get the value of the derivative.
    Last edited by mr fantastic; March 18th 2009 at 06:29 PM. Reason: Fixed a latex tag
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  6. #6
    Member OnMyWayToBeAMathProffesor's Avatar
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    i'm sorry but i fail to see how <br />
f(2x) - f(x) = 6x^2<br />
is given. a and b are two different integers and i did not think that you could just combine the 2. and where does 6x^2 come in? I am sorry but I think i am very lost.
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  7. #7
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    i'm sorry but i fail to see how <br />
f(2x) - f(x) = 6x^2<br />
is given. a and b are two different integers and i did not think that you could just combine the 2. and where does 6x^2 come in? I am sorry but I think i am very lost.
    You can let a and b be anything you want. They are merely pronumerals.

    Quote Originally Posted by mr fantastic View Post
    Let a = b = x: f(2x) - f(x) = 6x^2 \Rightarrow f(x) = 2x^2 + c where c is a constant.
    f(a + b) = f(x + x) = f(2x).

    f(a) = f(x)

    4ab + 2b^2 =  4x^2 + 2x^2 = 6x^2.

    And the result I posted follows.
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