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Math Help - Trying to learn how to do domain/range questions. Could you please help?

  1. #1
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    Trying to learn how to do domain/range questions. Could you please help?

    The question I have is:

    The function f has domain [1, \infty] and is defined by

    f(x) = ln (5x - 4) + 2.

    The question wants
    a] an expression for f^{-1}(x)
    b] statement of the domain and range of f^{-1}(x).

    I am trying to learn how to solve questions such as these, however I have no resources from which to learn it so it's very difficult not knowing how to do something without a lot of outside help!

    How would you solve this problem? If possible, is there any website that you could find that may help with this topic as I have tried looking and failed to find anything relating to it whatsoever
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  2. #2
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    How did you get [1,big]?

    You can't have "big]", it must be "big)".

    Tinker with: 5x-4 > 0
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    How did you get [1,big]?

    You can't have "big]", it must be "big)".

    Tinker with: 5x-4 > 0
    No no, it's [1, infinity]
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  4. #4
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    Quote Originally Posted by db5vry View Post
    The function f has domain [1, \infty] and is defined by
    f(x) = ln (5x - 4) + 2.
    The question wants
    a] an expression for f^{-1}(x)
    Here is [a].
    \begin{gathered}<br />
  x = \ln \left( {5y - 4} \right) + 2 \hfill \\<br />
  \ln \left( {5y - 4} \right) = x - 2 \hfill \\<br />
  5y - 4 = e^{x - 2}  \hfill \\ <br />
\end{gathered}
    Solve for y.
    Graph both to see if it works.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Here is [a].
    \begin{gathered}<br />
  x = \ln \left( {5y - 4} \right) + 2 \hfill \\<br />
  \ln \left( {5y - 4} \right) = x - 2 \hfill \\<br />
  5y - 4 = e^{x - 2}  \hfill \\ <br />
\end{gathered}
    Solve for y.
    Graph both to see if it works.
    Okay so my expression for f^{-1}(x) correctly is

    \frac{e^{y - 2} + 4}{5}

    How do you find the domain and range of this function, though?

    Thanks for the help so far
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  6. #6
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    Quote Originally Posted by db5vry View Post
    \frac{e^{y - 2} + 4}{5}
    How do you find the domain and range of this function, though?
    Do as I suggested. Graph the function.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by db5vry View Post
    Okay so my expression for f^{-1}(x) correctly is

    \frac{e^{y - 2} + 4}{5}

    How do you find the domain and range of this function, though?

    Thanks for the help so far
    Technically, you need to replace y with x for this to be the inverse function.

    Can you think of ANY number x such that
    y=\frac{e^{x - 2} + 4}{5} would be undefined? I can't. What do you think that means? Whta does this say about the domain?

    As far as the range is concerned, Can this function ever be negative? Does it keep get bigger as x gets bigger? What about if x gets smaller and smaller?

    3333
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  8. #8
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    Quote Originally Posted by db5vry View Post
    How did you get [1,big]?

    You can't have "big]", it must be "big)".

    Tinker with: 5x-4 > 0
    No no, it's [1, infinity]
    Unless this is sarcasm, you're missing the point. We don't put a bracket next to ∞ -- it's always a parenthesis. So the domain should be written as [1, ∞).

    Speaking of the domain, note that it is restricted. Under normal circumstances, the domain would be (0.8, ∞), because you can take a natural log of positive numbers only, and 5x - 4 > 0 when x > 0.8. Either the [place where this problem comes from] intentionally restricted the domain further, or it's a typo. I actually think it's the former.

    I would like to take a different approach to this problem. Instead of asking what the domain of the inverse is:
    f^{-1}(x) = \frac{e^{x - 2} + 4}{5}
    ask yourself, what is the range of f(x)?

    First consider this expression by itself:
    \ln (5x - 4)
    when you take natural logs of the numbers 1 and greater (since this is the domain of f(x)), you get non-negative numbers, or [0, ∞). But
    f(x)\;=\;\ln (5x - 4) + 2. That means the range is going to be [2, ∞), because you're taking all non-negative numbers from the natural log expression alone and then add 2.

    Remember the rule about domain/range of functions and their inverses. If the function is one-to-one, then the domain of the original is the range of the inverse, and the range of the original is the domain of the inverse.

    The domain is not (-∞, ∞) as VonNemo19 is suggesting, because we're not looking at this
    f^{-1}(x) = \frac{e^{x - 2} + 4}{5}
    by itself. In other words, under normal circumstances the domain would be (-∞, ∞), but since we're looking at this as an inverse of the function, we have to check the domain and range of the original function f(x)\;=\;\ln (5x - 4) + 2 and swap. The same holds true for the range of the inverse.

    So, for the inverse...
    Domain: [2, ∞)
    Range: [1, ∞)


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  9. #9
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    Quote Originally Posted by db5vry View Post
    No no, it's [1, infinity]
    Do you understand that a square bracket means that the end-point is included?

    It's [1, infinity), as has already been pointed out in post #2.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yeongil
    The domain is not (-∞, ∞) as VonNemo19 is suggesting, because we're not looking at this

    by itself. In other words, under normal circumstances the domain would be (-∞, ∞), but since we're looking at this as an inverse of the function, we have to check the domain and range of the original function and swap. The same holds true for the range of the inverse.
    I hate it that you are right, but I love it that you corrected me. Thanx.
    Last edited by mr fantastic; June 20th 2009 at 10:54 PM. Reason: Added quote tags.
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