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Math Help - Help woth functions

  1. #1
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    Help woth functions

    Hello!

    I need help to solve this:

    If F( ( x + 3 ) / ( x + 1 ) ) = 3x + 2 x belongs to R \ {!-1}
    then F( 5 ) = ?

    It is a first time ever for me to have met this kind of problems and functions in general (I'm trying to teach myself math to get on university), I have very weak background in math so you can imagine my struggle.

    I tried to do something like this:

    x = 1
    F( ( 1 + 3 ) / ( 1 + 1 ) ) = 3 * 1 + 2
    F( 4 / 2 ) = 5

    so, I have now 4/2 and 5 and I say F(4/2) = 5
    and to find F(5) I did this: first I found by how much I need to multiply 4/2 to get 5:
    4/2 * y = 5
    y = 5 * (2/4)
    y = 10/4
    y = 5/2

    now, I know that in order to get the value of F(5) I need to divide y with f(4/2):
    f(4/2) = 5
    so, y / 5 = 5/2 * 1/5 = 1/2 --- and the book states that is the right answer, BUT when I used for x some other number, like 3, I didn't get that answer. Please help.
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  2. #2
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    Solve \displaystyle \frac{x+3}{x+1} = 5
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  3. #3
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    They try to trick you with f(x)= bla bla.
    In general, it's helpful to read: f(x) as "f function of x".
    Hence, y(x) as "y function of x".

    So, in your case drop that F thing completely.
    (x+3)/(x+1) = some constant
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Solve \displaystyle \frac{x+3}{x+1} = 5
    Ugh... I don't get it...

    I solved and got 1/2, and thats the right answer, but...

    I mean, what's the purpose of the right side of equation then (3x + 2)?
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  5. #5
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    Quote Originally Posted by ikislav View Post
    Ugh... I don't get it...

    I solved and got 1/2, and thats the right answer, but...

    I mean, what's the purpose of the right side of equation then (3x + 2)?
    You are looking for F(5) . The number 5 tells you what x is, then the 3x + 2 gives you the value for F(5).

    -Dan

    If this is really giving you problems, there is a way to put F into a more standard form.

    Let \displaystyle y = \frac{x + 3}{x + 1}

    Solve for x:
    \displaystyle x = - \frac{y - 3}{y - 1}

    and put this in for F():
    \displaystyle F(y) = -3 \left ( \frac{y - 3}{y - 1} \right ) + 2

    and solve your problem by putting y = 5.
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