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16 September, 23:02

Let f = xy = 2x + y + 1 There is at least one x ∈ R with no y ∈ R. Find all such problematic x ∈ R. Then restrict the domain of f to exclude those problematic x and prove that this makes f a function.

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  1. 17 September, 01:42
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    Problematic x is x = 1

    Step-by-step explanation:

    Equation:

    xy = 2x + y + 1

    xy - y = 2x + 1

    y (x-1) = 2x + 1

    y = (2x+1) / (x-1)

    The problematic x is such that when the denominator of the function is 0

    x - 1 = 0

    x = 1 (the problematic x)

    So the domain of f is: x is the subset of R (real number) with the exception of x = / 1 (x not equal to 1)

    To prove this, we can plot the graph and in the graph we can see that as the value of x approaches from negative values to 1, y value will approaches negative infinity, and as the value of x approaches from large positive numbers, y value approaches infinity.

    In other words, we'll see an assymptote at x=1

    To prove that it is a function, we can do vertical line test by drawing vertical lines accross the graph. We'll see that each line crosses the equation line once hence proving the equation as a function
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