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Math Help - Differentiable function

  1. #1
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    Differentiable function

    Let f be a differentiable function, defined for all real numbers x, with the following properties. Find f(x). Show your work.
    - f '(x)=ax^2 + bx
    - f '(1)=6 and f ''(1)=18
    . 2
    - ∫ f(x)dx=18
    . 1

    I am completely lost on this question. I assume that I have to take the anti derivative of the first given property by using the second given to solve for the values of a and b, but I'm not sure how to do it.
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  2. #2
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    Quote Originally Posted by DarkestEvil View Post
    Let f be a differentiable function, defined for all real numbers x, with the following properties. Find f(x). Show your work.
    - f '(x)=ax^2 + bx
    - f '(1)=6 and f ''(1)=18
    . 2
    - ∫ f(x)dx=18
    . 1

    I am completely lost on this question. I assume that I have to take the anti derivative of the first given property by using the second given to solve for the values of a and b, but I'm not sure how to do it.
    f'(x) = ax^2 + bx

    f'(1) = a(1^2) + b(1) = a+b = 6

    f''(x) = 2ax + b

    f''(1) = 2a(1) + b = 2a + b = 18

    you have a system of equations in a and b ... solve for both.

    once you get a and b, integrate f'(x) to get f(x) with its constant of integration ... then use the information given by the definite integral.
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  3. #3
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    Quote Originally Posted by DarkestEvil View Post
    Let f be a differentiable function, defined for all real numbers x, with the following properties. Find f(x). Show your work.
    - f '(x)=ax^2 + bx Mr F says: Therefore {\color{red}f(x) = \frac{a}{3} x^3 + \frac{b}{2} x^2 + c}.

    - f '(1)=6 Mr F says: Therefore {\color{red}6 = a + b} .... (1)

    and f ''(1)=18 Mr F says: {\color{red}f''(x) = 2ax + b}. Therefore {\color{red}18 = 2a + b} .... (2)
    . 2
    - ∫ f(x)dx=18
    . 1
    Mr F says: Therefore {\color{red}18 = \left[\frac{a}{12} x^4 + \frac{b}{6} x^3 + cx + d\right]_1^2} .... (3)

    I am completely lost on this question. I assume that I have to take the anti derivative of the first given property by using the second given to solve for the values of a and b, but I'm not sure how to do it.
    Solve equations (1) and (2) simultaneously for a and b. Equation (3) gives a relationship between c and d but without further information exact values cannot be found.


    Edit: Re: "Equation (3) gives ....." etc. My mistake, d obviously cancels so getting c is simple.
    Last edited by mr fantastic; October 12th 2009 at 07:48 PM.
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    Wow that makes so much more sense now using system of equations. I found that a=12 and b=-6. So now I'm guessing I take the anti derivative of f'(x), making it f(x)=(ax^3/3) + (bx^2/2) + C. But wouldn't I have to solve for C to be able to find f(x)?
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    Quote Originally Posted by DarkestEvil View Post
    Wow that makes so much more sense now using system of equations. I found that a=12 and b=-6. So now I'm guessing I take the anti derivative of f'(x), making it f(x)=(ax^3/3) + (bx^2/2) + C. But wouldn't I have to solve for C to be able to find f(x)?
    that is correct ... use the given definite integral information to solve for C
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    Hmmm... how do I find C without an x value? Do I use f'(x) to find the value of x by plugging in a and b? And even so, how do I find C when I have both unknowns f(x) and C in the same equation of f(x)=(ax^3/3)+(bx^2/2)+C?
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  7. #7
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    Quote Originally Posted by DarkestEvil View Post
    Hmmm... how do I find C without an x value? Do I use f'(x) to find the value of x by plugging in a and b? And even so, how do I find C when I have both unknowns f(x) and C in the same equation of f(x)=(ax^3/3)+(bx^2/2)+C?
    evaluate and get an equation in terms of C, then solve for C ...

    \int_1^2 4x^3 - 3x^2 + C \, dx = 18
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  8. #8
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    Wow I can't believe I didn't see that.
    So through intense calculations, I end up with C=19 if I'm correct. So now I plug in the x values 1 and 2 and subtract and what not, right?
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