Online Course Discussion Forum
Math Challenge II-B
First note that when x=2‾√x=2, y=22‾√y=22, so the domain of the inverse will be all x≥22‾√x≥22.
We want to solve x=y+2yx=y+2y for yy. Clearing denominators gives xy=y2+2xy=y2+2 so y2−xy+2=0y2−xy+2=0. By the quadratic formula y=x±x2−8‾‾‾‾‾‾√2y=x±x2−82. For x≥2‾√x≥2 we want y=x+x2−8‾‾‾‾‾‾√2y=x+x2−82, answer choice C, as the inverse.
I can't figure out why we want y=x+x2−8‾‾‾‾‾‾√2y=x+x2−82, but not both +-For the discussion, it might help looking at the graphs a bit. See here for the functions on desmos: https://www.desmos.com/calculator/l3qvmedn75
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!
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