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Math Help - Functions...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Functions...

    f:R-->R. f(x)=-2x^2+12x-13

    What is Image of f(x)?

    second question:
    Divide the domain R to 2 intervals D1 and D2 to get two inverse functions in these intervals. Calculate these functions.
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  2. #2
    Member Nacho's Avatar
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    The image is R, because f hasnīt restriction, you can draw the grafh and you will see that Im (f) = R

    To find the inverse you must divide R in two intervalo, you must find de maximun or minimun, we say that this point is X, then:

    <br />
D_1  = \left( { - \infty ,x} \right) \wedge D_2  = \left( {x,\infty } \right)<br />
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  3. #3
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    I'm afraid Nacho is incorrect. He may be thinking of the "domain" rather than the "range" or image. Complete the square on that quadratic: f(x)=-2x^2+12x-13= -2(x^2+ 6x+ 9)- 13+ 18= -2(x-3)^2+ 6. Now it should be clear that f(x) is never larger than 6. And that should make it clear how to solve for x to find the two "inverse" functions.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    I don't got it...
    Where is your calculation of the two inverse functions? This is the main problem.

    And thank you..
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Ohh.. now it's a little bit clearer...
    Some questions:
    -How can I prove that f:R-->(inf,6] is not inverse?
    -What are the two intervals that I must take to prove that f is inverse.
    I mean: f_1-->(inf,6] and f_2-->(inf,6], which intervals should I take (D_1, D_2)?

    And thank you a lot!
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  6. #6
    Member Nacho's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    I'm afraid Nacho is incorrect. He may be thinking of the "domain" rather than the "range" or image.
    I'm sorry, HallsofIvy undertands my little confunsion

    Quote Originally Posted by Also sprach Zarathustra View Post
    Where is your calculation of the two inverse functions? This is the main problem.
    My teacher always say that he doesnīt like this method, but it works fine

    <br />
\begin{gathered}<br />
  y =  - 2x^2  + 12x - 13 \Leftrightarrow  - 2x^2  + 12x - 13 + y = 0 \hfill \\<br />
   \Rightarrow x = \frac{{ - 12 \pm \sqrt {12^2  - 4 \cdot  - 2 \cdot \left( { - 13 + y} \right)} }}<br />
{{2 \cdot  - 2}} = \frac{{ - 12 \pm \sqrt {144 + 8 \cdot \left( {y - 13} \right)} }}<br />
{{ - 2}} \hfill \\ <br />
\end{gathered} <br />

    If <br />
x \in \left( { - \infty ,6} \right) \Rightarrow x = \frac{{ - 12 - \sqrt {144 + 8 \cdot \left( {y - 13} \right)} }}<br />
{{ - 2}}<br />

    Now you changes the X for the Y

    Then your inverse is: <br />
f^{ - 1} :\left( { - \infty ,6} \right) \to \mathbb{R},{\text{ }}f^{ - 1} \left( x \right) = \frac{{ - 12 - \sqrt {144 + 8 \cdot \left( {x - 13} \right)} }}<br />
{{ - 2}}<br />

    To <br />
x \in \left( {6,\infty } \right)<br />
is analogous
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