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Thread: Investigate f(2x - 1) = x + 4

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    Junior Member B9766's Avatar
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    Investigate f(2x - 1) = x + 4

    I ran across this question and am not sure of the answer. Needless to say, I'm having trouble with conceptualizing all of the facets of functions.

    The question is: Investigate and understand f(2x - 1) = x + 4

    I get two different possible answers -


    1. Let z = 2x - 1
    2. x = (z + 1)/2; Evaluating for x
    3. f(z) = x + 4; from the original function statement
    4. f(z) = [(z + 1)/2] + 4; Substituting for x from step 2
    5. f(2) = [(2 + 1)/2] + 4; Evaluating with z = 2
    6. f(2) = 3/2 + 4
    7. f(2) = 3/2 + 8/2 = 11/2



    BUT



    1. Let g(x) = 2x-1
    2. f(g(x)) = x + 4; Composition of 2 functions
    3. g(2) = 2(2) - 1 = 3; Evaluating with g(2)
    4. f(g(2)) = 3 + 4 = 7


    No? What did I miss here? Am I misunderstanding a fundamental concept in functions?
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    Re: Investigate f(2x - 1) = x + 4

    Quote Originally Posted by B9766 View Post
    I ran across this question and am not sure of the answer. Needless to say, I'm having trouble with conceptualizing all of the facets of functions.
    The question is: Investigate and understand f(2x - 1) = x + 4[LIST=1][*]Let z = 2x - 1[*]x = (z + 1)/2; Evaluating for x
    Not quit sure what investigate means here. What you have done in the quote is correct.
    I would finish with:
    because $2\left(\dfrac{z+1}{2}\right)-1=z$ then $f(z)=\dfrac{z+9}{4}$
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    Junior Member B9766's Avatar
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    Re: Investigate f(2x - 1) = x + 4

    Plato: I'm confused by your reply. I showed 2 different approaches with 2 different answers. Both can't be correct.

    I didn't follow your "finish with" response at all. The first part is obvious z =z. But it's unclear how you transformed that into the second part that is an entirely 3rd different answer.

    I would really appreciate it if you would clarify this.
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    Re: Investigate f(2x - 1) = x + 4

    Quote Originally Posted by B9766 View Post
    Plato: I'm confused by your reply. I showed 2 different approaches with 2 different answers. Both can't be correct.
    I didn't follow your "finish with" response at all. The first part is obvious z =z. But it's unclear how you transformed that into the second part that is an entirely 3rd different answer.
    Well, I did say that I do not understand what you are asked to do.
    Here is a standard question. If $f(3x+7)=2-6x$ then what is $f(x)~?$
    Solution:
    Use a dummy variable, $z=3x+7$ so $x=\left(\dfrac{z-7}{3}\right)$.
    That means $ \begin{align*}f(z)&=2-6\left(\dfrac{z-7}{3}\right)\\&=2-3(z-7)\\&=2-2z+14\\&=16-2z \end{align*}$

    Since $z$ is a dummy we have $f(x)=16-2x$.
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    Re: Investigate f(2x - 1) = x + 4

    Quote Originally Posted by B9766 View Post
    Plato: I'm confused by your reply. I showed 2 different approaches with 2 different answers. Both can't be correct.

    I didn't follow your "finish with" response at all. The first part is obvious z =z. But it's unclear how you transformed that into the second part that is an entirely 3rd different answer.

    I would really appreciate it if you would clarify this.
    They are not different answers. Your two answers are answering two different questions.

    $f (g (x)) \neq f (x) $
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    Re: Investigate f(2x - 1) = x + 4

    Why are you trying to evaluate $f(2)$? You should be getting an expression for $f(x)$.
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    Re: Investigate f(2x - 1) = x + 4

    Quote Originally Posted by Plato View Post
    Well, I did say that I do not understand what you are asked to do.
    Here is a standard question. If $f(3x+7)=2-6x$ then what is $f(x)~?$
    Solution:
    Use a dummy variable, $z=3x+7$ so $x=\left(\dfrac{z-7}{3}\right)$.
    That means $ \begin{align*}f(z)&=2-6\left(\dfrac{z-7}{3}\right)\\&=2-3(z-7)\\&=2-2z+14\\&=16-2z \end{align*}$

    Since $z$ is a dummy we have $f(x)=16-2x$.
    I followed all of this up to the last line. Why is z a "dummy" in the last line? z = 3x + 7, right?
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    Re: Investigate f(2x - 1) = x + 4

    Quote Originally Posted by B9766 View Post
    I followed all of this up to the last line. Why is z a "dummy" in the last line? z = 3x + 7, right?
    $f(z)=16-2z , f(t)=16-2t$ and $g(x)=16-2x$ are all the same function only with different names.
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    Re: Investigate f(2x - 1) = x + 4

    I think that's a good indicator of my lack of knowledge.
    I understand that f(x) = 2x says, put an input value in as x (for example 2) and get an output = 2x (or in this case f(2) = 2 * 2 = 4).
    I also understand that if g(x) = 3x-1 and f(x) = 2x, then f(g(x)) = 2(3x-1) = 6x-2. Evaluating with an input of 2, f(g(2)) = 6(2)-2 = 10

    I think I'm having problems wrapping my mind around a function with a polynomial on both sides of the equation.
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    Re: Investigate f(2x - 1) = x + 4

    Thank you Plato. I understand what you're saying now. I think where my confusion arises, as I tried to relate to SlipEternal, is in reconciling the concept of a function as a program with inputs and outputs. It's easy to see f(x)=x+4 with "x+4" being the program, x being the input and f(x) being the output.

    If, in order to evaluate a function in the sense of it being a program, one first has to simplify it down to a single variable as you did above, I can accept that as simply a rule without further explanation. But for the life of me I can't imagine a real-life application where it would be necessary to address a function like that.

    Is this in fact the case?
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    Re: Investigate f(2x - 1) = x + 4

    Quote Originally Posted by B9766 View Post
    I think that's a good indicator of my lack of knowledge.
    I understand that f(x) = 2x says, put an input value in as x (for example 2) and get an output = 2x (or in this case f(2) = 2 * 2 = 4).
    I also understand that if g(x) = 3x-1 and f(x) = 2x, then f(g(x)) = 2(3x-1) = 6x-2. Evaluating with an input of 2, f(g(2)) = 6(2)-2 = 10

    I think I'm having problems wrapping my mind around a function with a polynomial on both sides of the equation.
    Suppose $f(x)=\color{red}{x^2-x}~\&~g(x)=\color{blue}{2x+1}$
    Then $f\circ g(t)=f(g(t))=\color{red}{(}2x+1\color{red}{)^2-(}2x+1\color{red}{)}$ and $g\circ f(t)=g(f(t))=2(x^2-x)+1$
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    Re: Investigate f(2x - 1) = x + 4

    Yes Plato. I'm not sure where the t's came from but am guessing they're typos. I think you meant to say:
    Suppose f(x)=x^2−x & g(x)=2x+1
    Then f∘g(x)=f(g(x))=(2x+1)^2−(2x+1) and g∘f(x)=g(f(x))=2(x^2−x)+1

    But in both cases the functions were expressed a monomials - f(x) and g(x). Where I get confused is when the function is expressed as a polynomial like f(x^2-5) or g(3x+4). In these cases, the concept of plugging a value in for the variable is not obvious in the way it's done nor in the outcome.

    If I have a table of x and y values where y = f(x) = 2x, then I know that plugging in the value x=5 will produce the output y = 10.

    If my function is f(x-4) = x^2, then, is my table of values expressed in terms of x-4 and y or x and y? Let's say I make x = 8, do I get f(8-4) = 8^2? I know that's wrong but somehow we violated the analogy of a computer with inputs and outputs.

    Part of this is that I can fathom why the function f(x) = x^2+4x+2 would be used in real life. I can't imagine why anyone would want to evaluate something like:
    f(x^4-6x^3+5x^2-12x+7) = 36x^2+60x-199

    Is there anything in the world of calculus or physics that would require such an evaluation?

    I'm sure I'm belaboring the point. But I really need to understand how this works because I know I'll need it in calculus.
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