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Symmetricity of exponential graphs

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose two exponential functions, f(x) and g(x) are symmetric with respect to x = 2.

    [tex]f(x)=a^{bx-1}[/tex]
    [tex]g(x)=a^{1-bx}[/tex]

    Prove f(2) = g(2)

    2. Relevant equations



    3. The attempt at a solution
    This isn't actually a problem. This is a property used to solve a different problem which I really had a hard time understanding. Why is this true?
     
    Last edited: Nov 13, 2008
  2. jcsd
  3. Nov 13, 2008 #2

    Mark44

    Staff: Mentor

    I might be completely misunderstanding what you're trying to say, but the functions you give are not symmetric with respect to the line x = 2, and furthermore, f(2) [tex]\neq[/tex] g(2). The two functions are reciprocals of one another, though.

    [tex]f(2) = a^{2b - 1}[/tex]
    [tex]g(2) = a^{1 - 2b} = a^{-(2b -1)} = \frac{1}{a^{2b - 1}} = \frac{1}{f(2)}[/tex]

    If the two functions were the mirror images of each other across the line x = 2, it would have to be the case that f(2 + x) = g(2 - x). For example, f(3) would have to be equal to g(1), which is not the case.

    Can you clarify what you're trying to do?
     
  4. Nov 13, 2008 #3
    Well, here's the whole problem:

    The two exponential functions

    [tex]f(x)=a^{bx-1}[/tex]
    [tex]g(x)=a^{1-bx}[/tex]

    Satisfy the following conditions

    a)Functions y = f(x) and y=g(x) are symmetric to the line x = 2 when graphed
    b)[tex]f(4) + g(4) = \frac {5}{2}[/tex]

    Find a and b.
     
  5. Nov 13, 2008 #4

    Dick

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    a^(bx-1) is not symmetric with respect to any x value unless either b=0 or a=1. a=1 doesn't work if f(4)+g(4)=5/2. Hence b=0. So a+a^(-1)=5/2. That's easy to solve. Is this problem as silly as that?
     
  6. Nov 14, 2008 #5
    Perhaps the question is intended to mean

    f(2-x)=g(x-2) for all x.
     
  7. Nov 14, 2008 #6

    Dick

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    So b(2-x)-1=1-b(x-2) for all x? About all I get out of that is 1=(-1).
     
  8. Nov 14, 2008 #7
    Sorry, I meant of course that the condition is

    f(2+x)=g(2-x) for all x.
     
  9. Nov 14, 2008 #8

    Dick

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    No, I'm sorry. I knew what you meant to say, but I didn't check that f(2-x)=g(x-2) was not the symmetry we were both thinking of. Yeah, that seems to give a reasonable answer for a and b. l46kok, look at how I tried to derive the value of b incorrectly and use borgwal's corrected condition on f and g to get b.
     
  10. Nov 25, 2008 #9
    Something is wrong, the answer is supposed to be

    a+b = 1 though
     
  11. Nov 25, 2008 #10
    I asked my professor and he told me that

    g(x) = f(4 - x), f(x) = g(4 - x), since x is symmetric to 2

    Now I'm even more confused.
     
  12. Nov 25, 2008 #11

    Dick

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    Why don't you try substituting that condition into the definitions of f(x) and g(x)?
     
  13. Nov 25, 2008 #12
    g(x) = f(4 - x), f(x) = g(4 - x), and f(x - 2) = g(x + 2) actually all say the same thing in a slightly different way.
     
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