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Math Help - Question about domains

  1. #1
    Member rowe's Avatar
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    Question about domains

    f(x) = \frac{2x+2}{x+1}

    f(x) = \frac{2(x+1)}{(x+1)}

    f(x) = g(x) = 2

    Now, f(x) in the first case has the domain \lbrace x \in \mathbb{R} | x \neq -1 \rbrace, but the second, g(x), has x \in \mathbb{R}. Just a bit confused as to why this is, isn't it the same function?

    I know that you can define your domain for whatever you wish, but if I algebraically manipulate a set function with a given domain, shouldn't that hold?
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by rowe View Post
    f(x) = \frac{2x+2}{x+1}

    f(x) = \frac{2(x+1)}{(x+1)}

    f(x) = g(x) = 2

    Now, f(x) in the first case has the domain \lbrace x \in \mathbb{R} | x \neq -1 \rbrace, but the second, g(x), has x \in \mathbb{R}. Just a bit confused as to why this is, isn't it the same function?

    I know that you can define your domain for whatever you wish, but if I algebraically manipulate a set function with a given domain, shouldn't that hold?
    Hi rowe,

    I assume the g(x) function is the f(x) factored, right?

    ] f(x) = \frac{2x+2}{x+1}

    g(x) = \frac{2(x+1)}{(x+1)}

    f(x) = g(x) = 2

    Remember the domain of f(x)=\frac{2x+2}{x+1} is all real numbers except x = -1 (which would make the denominator zero)

    When a value of x sets both the denominator and the numerator of a rational function equal to 0, there is a hole in the graph; that is, a single point at which the function has no value. g(x)=\frac{2(x+1)}{x+1} has a hole at x = -1.
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  3. #3
    Member rowe's Avatar
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    Hi, thanks.

    Let's just ignore g(x) for now.

    I'm aware that f(-1) is undefined, but what I'm saying is that I can algebraically manipulate f(x) = \frac{2x+2}{x+1} to make f(x) = 2 by the above process.

    The problem is that in doing so, it seems the domain of the function "changes". There is no x \in \mathbb{R} such that f(x)=2 is undefined.

    Or, do we have to keep the domain of f fixed, so even though I manipulate it, the domain stays the same so even though f(x)=2, we have said that we can't allow x to take the value -1?

    By that logic, I can algebraically manipulate f(x)=2 by:

    f(x)=2\cdot\frac{x+1}{x+1}

    f(x)=\frac{2(x+1)}{x+1}

    f(x)=\frac{2x+2}{x+1}

    But in this case, I could have orignally defined the domain of f for all real numbers! In both cases, the domain changes. What's the story?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by rowe View Post
    Hi, thanks.

    Let's just ignore g(x) for now.

    I'm aware that f(-1) is undefined, but what I'm saying is that I can algebraically manipulate f(x) = \frac{2x+2}{x+1} to make f(x) = 2 by the above process.

    The problem is that in doing so, it seems the domain of the function "changes". There is no x \in \mathbb{R} such that f(x)=2 is undefined.

    Or, do we have to keep the domain of f fixed, so even though I manipulate it, the domain stays the same so even though f(x)=2, we have said that we can't allow x to take the value -1?

    By that logic, I can algebraically manipulate f(x)=2 by:

    f(x)=2\cdot\frac{x+1}{x+1}

    f(x)=\frac{2(x+1)}{x+1}

    f(x)=\frac{2x+2}{x+1}

    But in this case, I could have orignally defined the domain of f for all real numbers! In both cases, the domain changes. What's the story?
    I'm not sure what your confusion is. Let me state it clearly.

    At x = -1, f(x) is undefined. That means there's a hole in the graph at (-1, 2).

    F(x) = 2 for all real numbers in the domain except -1.

    The graph will be a horizontal line through y = 2 with a point of discontinuity at (-1, 2).
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  5. #5
    Member rowe's Avatar
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    I'm not too concerned with where the discontinuity is, I'm asking if I have a function f:R \to R, with f(x)=2, and I manipulate it so that f(x)=\frac{2x+2}{x+1}, is this an "illegal" move so to speak, because the domain doesn't support the discontinuity?

    And the same vice versa- the domain inherently denoted for all real x, bar -1 in f(x)=\frac{2x+2}{x+1}, that won't change the domain to all R when we change f(x)=2?

    Don't worry too much about it, from what you're saying and what I think I've pretty much convinced myself you have to stick to the domain given by the function definition.
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