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Math Help - Functions

  1. #1
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    Functions

    Hi all!

    Just having a little problem with the following:

    Functions f and g are defined as follows

    <br />
f(x) = 2x + 3, x\epsilon\Re and g(x) = \frac{1}{x-1}, x\epsilon\Re , x [does not equal to] 1

    I couldn't find the correct symbol for not equal to sign.

    (i) Find the expression for the inverse function f^-1 (x)

    (ii) Find the expression for the composite function gf (x)

    (iii) Solve the equation f^-1 (x) = gf(x)-1

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by gary223 View Post
    Hi all!

    Just having a little problem with the following:

    Functions f and g are defined as follows

    <br />
f(x) = 2x + 3, x\epsilon\Re and g(x) = \frac{1}{x-1}, x\epsilon\Re , x [does not equal to] 1

    I couldn't find the correct symbol for not equal to sign.
    \ne "\ne".

    (i) Find the expression for the inverse function f^-1 (x)
    If y= f(x), then x= f^{-1}(y). If y= 2x+3, solve for x, then "swap" x and y so you can write it as f^{-1}(x) rather than f^{-1}(y).

    (ii) Find the expression for the composite function gf (x)
    g(f(x))= g(2x+3). Replace "x" by "2x+3" in the formula for g.

    (iii) Solve the equation f^-1 (x) = gf(x)-1

    Thanks in advance.
    Use your answers to (i) and (ii) to get the equation for (iii).
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    \ne "\ne".


    If y= f(x), then x= f^{-1}(y). If y= 2x+3, solve for x, then "swap" x and y so you can write it as f^{-1}(x) rather than f^{-1}(y).


    g(f(x))= g(2x+3). Replace "x" by "2x+3" in the formula for g.


    Use your answers to (i) and (ii) to get the equation for (iii).
    Thanks for the helpful reply but I'm still having problems actually solving the three parts. Could you post the answer so I can work from that example when trying to do other questions I have similar to this.
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  4. #4
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    Quote Originally Posted by gary223 View Post
    Thanks for the helpful reply but I'm still having problems actually solving the three parts. Could you post the answer so I can work from that example when trying to do other questions I have similar to this.
    How about you post what you have done?

    In the first problem, I suggested that you solve y= 2x+3 for x. Did you do that?

    In the second, I told you to replace x by 2x+ 3 in the formula for g(x). Since that formula was g(x)= \frac{1}}{x-1}, replacing x by 2x-3 gives g(f(x))= g(2x+3)= \frac{1}{(2x+3)-1}. What is that?
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  5. #5
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    Here's what I got for the first part:

     y = 2x+3
     y - 3 = 2x
     \frac{y-3}{2} = x

    Correct?
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  6. #6
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    And I got g(f(x)) = \frac{1}{2x+2}

    Please help.
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  7. #7
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    Quote Originally Posted by gary223 View Post
    Here's what I got for the first part:

     y = 2x+3
     y - 3 = 2x
     \frac{y-3}{2} = x

    Correct?
    Very good- so far! Now answer the question- what is f^{-1}(x)?
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  8. #8
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    Quote Originally Posted by gary223 View Post
    And I got g(f(x)) = \frac{1}{2x+2}

    Please help.
    Yes, that is correct.
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  9. #9
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    For iii)

    I got:

    \frac{y-3}{2} = \frac{1}{2x+2} - 1

    =\frac{y-3+1}{2} = \frac{1}{2x+2}

    =\frac{y-2}{2} = \frac{1}{2x+2}

    Is this correct? If not, what is the rest?
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  10. #10
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    Quote Originally Posted by gary223 View Post
    Here's what I got for the first part:

     y = 2x+3
     y - 3 = 2x
     \frac{y-3}{2} = x

    Correct?
    Quote Originally Posted by HallsofIvy View Post
    Very good- so far! Now answer the question- what is f^{-1}(x)?
    Quote Originally Posted by gary223 View Post


    For iii)

    I got:

    \frac{y-3}{2} = \frac{1}{2x+2} - 1

    =\frac{y-3+1}{2} = \frac{1}{2x+2}

    =\frac{y-2}{2} = \frac{1}{2x+2}

    Is this correct? If not, what is the rest?
    I told you before that you had not finished problem 1. You got  \frac{y-3}{2} = x But the problem was to find f^{-1}(x), not f^{-1}(y)! You need to write "x" in place of "y" to make that a function of x.

    That's why you have an equation involving both x and y when you should have an equation in x only. Problem iii was to solve f^{-1}(x)= f(g(x))- 1, an equation in x only, for x.
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    I told you before that you had not finished problem 1. You got  \frac{y-3}{2} = x But the problem was to find f^{-1}(x), not f^{-1}(y)! You need to write "x" in place of "y" to make that a function of x.

    That's why you have an equation involving both x and y when you should have an equation in x only. Problem iii was to solve f^{-1}(x)= f(g(x))- 1, an equation in x only, for x.
    So its  \frac{x-3}{2} = y ?
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