Online Course Discussion Forum

Math Challenge II-B

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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!