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Math Help - Inverse Functions

  1. #1
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    Inverse Functions

    Will you please find domain and range of two functions: f(x) = 3x /(x + 2) and
    f (x) = -2x / (x-3) , which is the inverse of the first function.
    Thank you
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Will you please find domain and range of two functions: f(x) = 3x /(x + 2) and
    f (x) = -2x / (x-3) , which is the inverse of the first function.
    Thank you
    the domain of the function is the range of the inverse function and the range of the function is the domain for the inverse function.

    so find the domain of both functions and you should have everything.

    the domain is the set of all x-values for which the function is defined. so find what x's we CANNOT have, and then the domain is all x's but those. (Hint: what can we not do with a fraction?)

    note, the range is the set of y-values for which the function is defined. so when stating the range, make sure to do it in terms of y
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    Jhevon,

    Thank you for taking time and explaining all of this. I have solved the problem before I posted a question, but my answer does not coincide with the answer in the textbook.

    f(x) = 3x /(x + 2) Domain: All real numbers except x = -2 ; Range: All real numbers except y = 3 ( my reasoning: y = 3 is a horizontal asymptote and the function does not cross the horizontal asymptote, because when I tried to solve 3x /(x + 2) = 3 , I got 3x = 3x + 6, which does not make sense

    f(x) = -2x / (x-3) Domain: All real numbers except x = 3 ; Range: All real numbers except y = -2 as y = -2 is a horizontal asymptote of f(x) = -2x / (x-3) and does not cross the function.

    Is this correct? The textbook has a different answer for the range, I do not have it with me right now, I will post it later, but will you please check whether my answers are correct.

    Thank you
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Jhevon,

    Thank you for taking time and explaining all of this. I have solved the problem before I posted a question, but my answer does not coincide with the answer in the textbook.

    f(x) = 3x /(x + 2) Domain: All real numbers except x = -2 ; Range: All real numbers except y = 3 ( my reasoning: y = 3 is a horizontal asymptote and the function does not cross the horizontal asymptote, because when I tried to solve 3x /(x + 2) = 3 , I got 3x = 3x + 6, which does not make sense

    f(x) = -2x / (x-3) Domain: All real numbers except x = 3 ; Range: All real numbers except y = -2 as y = -2 is a horizontal asymptote of f(x) = -2x / (x-3) and does not cross the function.

    Is this correct? The textbook has a different answer for the range, I do not have it with me right now, I will post it later, but will you please check whether my answers are correct.

    Thank you
    that is correct. do you notice that the domain of the first is the range of the second (except we use x when we had y) and vice versa?

    it seems to me that you just tried to pick the horizontal asymptotes out of the blue. you could have found them directly using limits. but good stuff
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  5. #5
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    Inverse functions

    How could I pick up horizontal asymptotes out of the blue correctly? To find the horizontal asymptotes for the rational functions with the same power of x in the nominator and the denominator, I used the coefficients of x in the nominator and the denominator, so for the function f(x) = 3x /(x + 2) the horizontal asymptote is y = 3/1, and for the function f(x) = -2x / (x-3) the horizontal asymptote is y = 2/1.

    Does it make any sense?

    Thank you again.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    How could I pick up horizontal asymptotes out of the blue correctly? To find the horizontal asymptotes for the rational functions with the same power of x in the nominator and the denominator, I used the coefficients of x in the nominator and the denominator, so for the function f(x) = 3x /(x + 2) the horizontal asymptote is y = 3/1, and for the function f(x) = -2x / (x-3) the horizontal asymptote is y = 2/1.

    Does it make any sense?

    Thank you again.
    yes, that makes sense. but that's not what you said before. you said you noticed there was no solution to ... = 3 so 3 was a horizontal asymptote or whatever. but you are correct. i guess because you're in pre-calculus you don't know why you take the ratio of the coefficients, it is because you use limits. but don't worry about that. good job
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  7. #7
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    Jhevon,
    The answer from the textbook:
    Domain f: All real numbers except -2
    Range f = Domain f-1 : All real numbers except 3
    Why is the range of f-1 missing from the answer? Is it just a misprint or is there a reason for it?

    Thank you.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Jhevon,
    The answer from the textbook:
    Domain f: All real numbers except -2
    Range f = Domain f-1 : All real numbers except 3
    Why is the range of f-1 missing from the answer? Is it just a misprint or is there a reason for it?

    Thank you.
    maybe they just wanted the domain and range of the first function. which is what they gave you. anyway, if they did leave part of the answer out on purpose, the reason would be that it is implied. since we would know that domain of f^{-1} = range of f and the range of f^{-1} = domain of f, and they gave us those answers already
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    Jhevon,

    Thank you so very much.
    Have a great day!
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