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Math Help - Is my answer right? Function problem

  1. #1
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    Is my answer right? Function problem

    Find f(x) if f(x+1) = x^2-5x+3. Hint: let u =x+1 and find f(u). (ps do you think the author of this question is curseing at me? haha)

    I got f(x)= x(x+2)-9x because f(x+1) = (x+1)(x+1+2)-9x=x^2-5x+3.

    But this just seems fishy. like there can be other answers right? So am I right or am i missing something?
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  2. #2
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    \displaystyle f(x+1) = x^2 - 5x + 3

    So \displaystyle f(x) = f(x + 1 - 1) = (x - 1)^2 - 5(x - 1) + 3. Now simplify.
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    Quote Originally Posted by Prove It View Post
    \displaystyle f(x+1) = x^2 - 5x + 3

    So \displaystyle f(x) = f(x + 1 - 1) = (x - 1)^2 - 5(x - 1) + 3. Now simplify.
    I kind of understand what your doing... if f(x+1) = x^2-5x+3, then f(x-1) should equal f(x). So your putting x-1 into thef(x) equation that has f( x+1) equation in it.

    However.... (x-1)^2-5(x-1)+3 simplifys to x^2-7x+9.

    So your saying f(x) = x^2-7x+9

    But then how come f(x+1) = x^2-5x+10 ? shouldnt putting (x+1) threw f(x) = x^2-5x+3 like the original equation? thats why I dont get it.
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  4. #4
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    You were given the hint "let u =x+1 and find f(u)". Doing that, x= u-1 and f(u)= (u- 1)-^2- 5(u- 1)+ 3= u^2- 2u+ 1- 5u+ 5+ 3= u^2- 7u+ 9 which is the same as f(x)= x^2- 7x+ 9.

    Do you understand what "f(x)= x^2- 7x+ 9" means?

    What would f(a) equal? What about f(a+1)? What would f(x+ 1) equal?
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    Quote Originally Posted by HallsofIvy View Post
    You were given the hint "let u =x+1 and find f(u)". Doing that, x= u-1 and f(u)= (u- 1)-^2- 5(u- 1)+ 3= u^2- 2u+ 1- 5u+ 5+ 3= u^2- 7u+ 9 which is the same as f(x)= x^2- 7x+ 9.

    Do you understand what "f(x)= x^2- 7x+ 9" means?

    What would f(a) equal? What about f(a+1)? What would f(x+ 1) equal?
    No I dont understand. I thought that if f(x+1) equals the above equation, then if you put (x-1) threw it it might go back to f(x). Im pretty sure I was wrong thinking that. SO I dont really get what to do. (other then just make up any equation that when you put x+1 threw would equal the same as f(x+1)).

    As for x(a) and x(a+1), I dont know how to answer that without having f(x) to plug them into. I guess you MIGHT mean the constant before x^2 for "a".

    Please help.

    I dont really see where to go because im thinking it says find f(u), u = x+1. So thats really saying find f(x+1) isnt it? and isnt f(x+1) the equation that they give me? So thats not really helping me. So im definitly missing something.
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  6. #6
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    Quote Originally Posted by frankinaround View Post
    No I dont understand. I thought that if f(x+1) equals the above equation, then if you put (x-1) threw it it might go back to f(x). Im pretty sure I was wrong thinking that. SO I dont really get what to do. (other then just make up any equation that when you put x+1 threw would equal the same as f(x+1)).

    As for x(a) and x(a+1), I dont know how to answer that without having f(x) to plug them into. I guess you MIGHT mean the constant before x^2 for "a".

    Please help.

    I dont really see where to go because im thinking it says find f(u), u = x+1. So thats really saying find f(x+1) isnt it? and isnt f(x+1) the equation that they give me? So thats not really helping me. So im definitly missing something.
    Here's another way to look at it.
    Then you could re-examine the previous posts by "Prove It" and "HallsoIvy"
    since the logic is more compact and faster.

    f(x+1)=x^2-5x+3

    We can try to write this in terms of (x+1) instead of x.

    (x+1)^2=(x+1)(x+1)=x^2+2x+1

    \Rightarrow\ x^2=(x+1)^2-2x-1

    We can also write the "x" term as an (x+1) term...

    x^2-5x+3=(x+1)^2-2x-1-5x+3=(x+1)^2-7x+2

    -7x+2=-7(x+1)+7+2

    \Rightarrow\ x^2-5x+3=(x+1)^2-7(x+1)+9

    These equations are identical,

    but they are both f(x+1)

    Hence, to write f(x), we can now substitute x for x+1
    in the equation written in terms of (x+1).
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  7. #7
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    Quote Originally Posted by frankinaround View Post
    No I dont understand.
    Then you really need to go over "function notation" with your teacher.

    I thought that if f(x+1) equals the above equation, then if you put (x-1) threw it it might go back to f(x).
    No, I didn't. I replaced x with u- 1 which is equal to x.

    Im pretty sure I was wrong thinking that. SO I dont really get what to do. (other then just make up any equation that when you put x+1 threw would equal the same as f(x+1)).

    As for x(a) and x(a+1), I dont know how to answer that without having f(x) to plug them into. I guess you MIGHT mean the constant before x^2 for "a".

    Please help.

    I dont really see where to go because im thinking it says find f(u), u = x+1. So thats really saying find f(x+1) isnt it? and isnt f(x+1) the equation that they give me? So thats not really helping me. So im definitly missing something.
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  8. #8
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    Hello, frankinaround!

    This is a variation of what Archie did . . .


    \text{Find }f(x)\,\text{ if }f(x+1) \:=\: x^2-5x+3.

    With a little imagination, we can hammer the function into the desired form.


    We have: . f(x+1) \;=\;x^2 - 5x + 3


    Add and subtract (2x+1)

    . . f(x+1) \;=\;x^2 {\bf + 2x + 1} - 5x + 3 \:{\bf -\,2x - 1}

    . . . . . . . . . \;=\;(x+1)^2 - 7x + 2


    Subtract and add 7:

    . . f(x+1) \;=\;(x+1)^2 -7x {\bf- 7} + 2 \:{\bf+\: 7}

    . . . . . . . . . =\;(x+1)^2 - 7(x+1) + 9


    Therefore: . f(x) \;=\;x^2 - 7x + 9


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I just found yet another approach.


    Let the original function be: . f(x) \;=\;Ax^2 + Bx + C


    Since f(x+1) \:=\:x^2 - 5x + 3, we have:

    . . A(x+1)^2 + B(x+1) + C \;=\;x^2 - 5x + 3


    This simplifies to:

    . . Ax^2 + (2A+B)x + (A+B+C) \;=\;x^2-5x+3


    Equate coefficients: . \begin{Bmatrix}A \;=\;1 \\ 2A + B \;=\;\text{-}5 \\ A+B+C \;=\;3 \end{Bmatrix}

    . . and solve the system: . A = 1,\;B = \text{-}7,\;C = 9


    Therefore: . f(x) \;=\;x^2 - 7x + 9

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