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Math Help - functions

  1. #1
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    functions

    Question:

    a) f(x) = | (3-x)/(x-2) |, x > 2

    Define the inverse funtion of f corresponding to the domain of f.

    b) g(x) = ln (x+2), x > -2
    h(x) = 2 + (e ^ -x) , x > b and b > 0

    The composite function gh is well defined and the range of gh is given as (ln 4, ln 6]. Find the exact value of b.

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  2. #2
    MHF Contributor red_dog's Avatar
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    a)
    I think f has not an inverse, because f is not one-to-one function:
    \displaystyle f\left(\frac{8}{3}\right)=f(4)=\frac{1}{2}
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  3. #3
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    Quote Originally Posted by Tangera View Post
    Question:

    a) f(x) = | (3-x)/(x-2) |, x > 2

    Define the inverse funtion of f corresponding to the domain of f.

    ...
    1. Write the equation of f as a piecewise defined function: (completely drawn in black)

    f(x)=\left\{\begin{array}{lcl}\dfrac{x-3}{2-x}~,& \text{for} & 2 \leq x < 3 \\ \\<br />
\dfrac{x-3}{x-2}~,& \text{for} &  x \geq 3 \end{array}\right.

    2. Since f(3) = 0 this is the minimum of f. The behavior of the function changes at x = 3 from monotically decreasing to monotically increasing.

    3. Therefore the inverse function must be defined piecewise too:

    f^{-1}(x)=\left\{\begin{array}{lcl}\dfrac{2x+3}{x+1}~,  & \text{for} &  x > 3 \\ \\<br />
\dfrac{2x-3}{x-1}~,& \text{for} &  0 \leq x \leq1 \end{array}\right.

    EDIT: As red_dog pointed out there doesn't exist one inverse function. Therefore you have to split my last answer into:

    f_1^{-1}(x)= \frac{2x+3}{x+1}~, \text{for }   x > 0 and

    f_2^{-1}(x)= \frac{2x-3}{x-1}~, \text{for }   x > 0
    Attached Thumbnails Attached Thumbnails functions-betrag_umkehrfkt.gif  
    Last edited by earboth; June 21st 2008 at 12:11 AM.
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