Question

Let f = {(x, y) ∈ R × R | 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.

Answer 1

Answer:

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