I want proof that the following function is strictly monotone (I need this proof to prove something else).
The problem is I may not use the first derivate because we have not discussed that in our readings yet.
It would make the work way easier but I'm not allowed to do it…
$f(x) = x + \sqrt[5]{x} – \frac{1}{x}$
We have said that $f$ is monotonically decreasing (or monotonically increasing) if for all $x_{1}, x_{2} \in M$ with $x_{1} \leq x_{2}: f(x_{1}) \leq f(x_{2})$ (or $f(x_{1}) \geq f(x_{2})$).
But we were not told what $x_{1}$ and $x_{2}$ stand for so I don't really know what to do with this formula.
It's really annoying we cannot just use the derivative 🙁
I could prove it I think if I knew what these two variables stand for, what I have to put into the function for these two variables?
Best Answer
What it means is that for any two numbers $x_1<x_2$ in the domain of the function you have $f(x_1) < f(x_2)$. If you want to use this (definition) to prove that your function is strictly increasing, then you need to do this for all $x_1 < x_2$. So let's say that we had two such numbers.
Then (you probably already know this) $x_1 < x_2$ and $\sqrt[5]{x_1} < \sqrt[5]{x_2}$ because both $x$ and $\sqrt[5]{x}$ are increasing functions.
Now then, if $x_1 < x_2$ then $\frac{1}{x_1} > \frac{1}{x_2}$. That means $-\frac{1}{x_1} < - \frac{1}{x_2}$ (at least for $x_1, x_2$ with the same sign.
Since all pieces have increased, you have $f(x_x) < f(x_2)$.