Infimum of $2^x + 2^{1-x}, x \in \mathbb{R}$. Inequality of means? Interesting points on quadratic function.

Question

So I checked some exercises in my book and found 2 interesting ones. It is required to determine infimum of sets given by:

$$\frac{x}{x^2+1}, x \in \mathbb{R}$$ $$2^x + 2^{1-x}, x \in \mathbb{R}$$

The interesting thing about those was that usually the numbers are naturals and here I got $x \in \mathbb{R}$. Because of that I have no boundary on left side of Xes. I wasn't able to transform those in any meaningful way, aside from: $$2^x + 2^{1-x} = 2^x + \frac{2}{2^x} = \frac{2^x*2^x +2}{2^x} \implies 2^x = t \implies \frac{t^2 +2}{t}$$

Then I saw that both functions describing both sets are alike: quadratic function in numerator, linear function in denominator. I started to play with those functions in desmos plotter and got an interesting information. There is a point in quadratic function where quadratic function in no longer growing "slower" than linear function and starts to grow faster. That is obviously the point of infimum. By that time I already got the answer for the first one - inequality of means. And here are my questions:

• if that point exists for every quadratic function, how it's called (so I can read some more about it)?
• how is the growth of quadratic function past that point compared to "traditional" exponential - they seem similar on picture?
• is it possible to say that the inf of second set is where the infimum of expression with variable $t$ is? It stands for ever-positive variable so it would look like it can be done by inequality of means too

Answer 1

Both of those expressions have a more important link, and that's symmetry. Consider the following argument for the second one: suppose a point $x_0$ was an extrema for the function. However notice that

$$2^{x_0}+2^{1-x_0} = 2^{1-x_0}+2^{x_0}$$

which is the point $1-x_0$. Since two distinct points evaluate to the same value on a continuous function, this implies the function has some sort of a bowl shape, and these points cannot be extrema, unless they weren't distinct. This leads the $x_0 = \frac{1}{2}$ and taking limits $x\to\pm\infty$ establishes it as a minimum.

The same argument applies for the first one, only the symmetry is $x_0 \leftrightarrow \frac{1}{x_0}$, which leads to the point $x_0 = 1$ being a maximum.