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Math Challenge II-B

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Math Challenge II-B
by zitian li - Wednesday, June 8, 2022, 3:03 PM

First note that when x=2x=2y=22y=22, so the domain of the inverse will be all x22x≥22.

We want to solve x=y+2yx=y+2y for yy. Clearing denominators gives xy=y2+2xy=y2+2 so y2xy+2=0y2−xy+2=0. By the quadratic formula y=x±x282y=x±x2−82. For x2x≥2 we want y=x+x282y=x+x2−82, answer choice C, as the inverse.

I can't figure out why we want y=x+x282y=x+x2−82, but not both +- 
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Re: Math Challenge II-B
by John Lensmire - Thursday, June 9, 2022, 2:21 PM

For the discussion, it might help looking at the graphs a bit. See here for the functions on desmos:

Let's start with the question itself. Why do we need $x\geq \sqrt{2}$ in the first place. Considering the graph of $y = x + 2/x$, $(\sqrt{2}, 2\sqrt{2})$ is the minimum, and if we don't restrict the domain, the function isn't invertible. (For example, if y = 3 then x is either 1 OR 2).

When we solve for y in the inverse we get $y = \dfrac{x \pm \sqrt{x^2-8}}{2}$ as mentioned, however, this is NOT a function. Looking at the graphs we see the "+" portion (graphed in blue) corresponds to when $x\geq \sqrt{2}$ in the original function (graphed in red) and the "-" portion (graphed in green) corresponds to when $x\leq \sqrt{2}$ in the original function.

Hope this helps!