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Math Help - find domain and assymptote of functions

  1. #1
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    find domain and assymptote of functions

    Consider the function f(x) = 2x + 9/7x - 11 and do the following:

    a) Find D of f
    B) Determine the equation of the vertical assmptote
    c) Solve for the independent variable in terms of the dependent variable
    d) Determine the equaton of the vertical assymptote of the graph of f(x)
    e) is f(x) on eo one? Explain using a graph of f(x).
    F) Find R of f
    G) Find a formual for the inverse function.
    H) Find the domain and range of the inverse function.
    i) Show a precise graph of f(x)
    j) Show a precise graph of the inverse.
    k) find (f of inverse)(x)
    l) find (inverse of f)(x)



    THANKS I APRREACIATE YOUR HELP!!!
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    Quote Originally Posted by rlarach View Post
    Consider the function f(x) = 2x + 9/7x - 11 and do the following:

    a) Find D of f
    B) Determine the equation of the vertical assmptote
    c) Solve for the independent variable in terms of the dependent variable
    d) Determine the equaton of the vertical assymptote of the graph of f(x)
    e) is f(x) on eo one? Explain using a graph of f(x).
    F) Find R of f
    G) Find a formual for the inverse function.
    H) Find the domain and range of the inverse function.
    i) Show a precise graph of f(x)
    j) Show a precise graph of the inverse.
    k) find (f of inverse)(x)
    l) find (inverse of f)(x)



    THANKS I APRREACIATE YOUR HELP!!!
    First of all, use parenthesis. I'm sure you mean
    f(x) = \frac{2x + 9}{7x - 11}
    not
    f(x) = 2x + \frac{9}{7x} - 11.

    a) Use the quotient rule. For f(x) = \frac{g(x)}{h(x)},
    f^{\prime}(x) = \frac{g^{\prime}(x)h(x) - g(x) h^{\prime}(x)}{h^2(x)}

    b) Vertical asymptotes typically show up where the denominator of the expression is 0.

    c) If you are talking about asymptotes in class you should already know how to do this.

    Call y = f(x):
    y = \frac{2x + 9}{7x - 11}

    (7x - 11)y = 2x + 9

    7xy - 11y = 2x + 9

    7xy - 2x = 11y + 9

    x(7y - 2) = 11y + 9

    x = \frac{11y + 9}{7y - 2}

    d) This is the same as question b)????

    e) Graph it. What is the test for a 1 to 1 function? (Hint: look up "horizontal line test."

    f) Look at your graph. What does it look like the range might be? Now for the harder question: Is there any value of y that cannot be reached by any x? (Hint: think about the horizontal asymptote. Can the function take on this value of y?)

    g) I did most of the work on this in c). You find the inverse function by switching the roles of x and y in the function and solving for the new y.

    h) The domain of the inverse function is the range of the function. The range of the inverse function is the domain of the function. (You can also look at the graph of the inverse function to find out what these are.)

    i) You should have already done this by now. If you haven't, do it now. You know the vertical and horizontal asymptotes. Now figure out where the function is positive and negative.

    j) Same as the process for i).

    k) "find (f of inverse)(x)" Huh?

    l) You already did this. f^{-1}(x) is your inverse function, just with the y replaced by f^{-1}(x). It's just a notation change.

    -Dan
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