Q:

Checking a fuction for monotony

I have a question about checking for monotony in functions and later proving them mathematically.
$$f_1:\mathbb {R}^+→\mathbb {R},x \to \frac{1+6x+2x^2}{(x+3)x} $$
By plugging in values 0, 1 and 2 I get the following results: 1, 2.25 and 2.05. I see this is no monotony so I don't have to explain my solution. (Or I am incorrect?)
$$f_1:\mathbb {R}→\mathbb {R},x \to \ln(x) $$
I plugged values 3,5 and 7 here, resulting in 1.09, 1.60 and 1.94, thus making this a monotonically increasing function. 
My question is how can I prove this mathematically? I already know to confirm this is as a monotonically increasing function, the following statements have to be correct: when for $x_1<x_2$ applies $f(x_1)<f(x_2)$. I just don't understand which values I use to prove this.

A:

You say that $f: D \to \mathbb{R}$ is (strictly) increasing if and only if for all $x_1,x_2 \in D$ you have that $x_1 < x_2 \Rightarrow f(x_1)<f(x_2)$. So, if you want to prove that $f$ is not increasing, you just need to find a counterexample (if it was increasing, the implication   should hold for every $x_1,x_2$. However, if you want to prove that the function is increasing, an example is not enough... because you must show that the implication holds for all $x_1,x_2$.
You cannot prove that all dogs are brown just by giving an example of a brown dog, but you can show that not all dogs are brown by finding a black dog.
For instance, regarding the logarithm, taking $0 < x_1< x_2$, you can note that
$$
f(x_1)-f(x_2) = \ln x_1-\ln x_2 = \ln \frac{x_1}{x_2} < 0 \Rightarrow f(x_1) < f(x_2).
$$
The  inequality $\ln \frac{x_1}{x_2}< 0$ is obtained because $0 < x_1/x_2 < 1$ and so its logarithm is negative. Of course, from the previous is equivalent to $f(x_1)- f(x_2)$.

