# Ascertaining whether a function is increasing or decreasing

In a given question I'm given the following function

$$\int^{2x}_{x}\frac{1}{t}dt$$

The question asks if the function is increasing or decreasing on the interval $$(0,\infty)$$

I've taken the integral of the function and got $$\ln2$$ but I'm not sure how to decipher whether the function is increasing or decreasing given this information.

Any tips?

Well, the indefinite integral of $$\frac{1}{t}$$ is $$\ln t+C$$, so: $$f(x)= \int_x^{2x}\frac{1}{t}dt= \ln(2x)-\ln(x)= \ln\left(\frac{2x}{x}\right)= \ln(2)$$ Therefore, $$f$$ is constant so it is both decreasing and increasing.
• does this mean that on the interval $(0,\infty)$ the function is increasing? what do you mean when you say it is both increasing and decreasing when the integral is a constant? – esc1234 Apr 20 at 13:56
• @esc1234 $f$ is constant in that interval. If you examine the definition, you’ll see that a constant function is both increasing and decreasing. – Yuval Gat Apr 20 at 14:01
• i marked your answer as the answer to my original question. The key part of the answer for me was that "$f$ is constant so it is both decreasing and increasing. But I am still not sure where this information is coming from. What definition are you referring to that states if the integral of a function is a constant it is both increasing and decreasing? thanks for your feedback btw – esc1234 Apr 20 at 20:58
• @esc1234 you need to forget about the integral. $f$ is defined by an integral, but after evaluating we saw that it is identically equal to $\ln 2$. The definition of an increasing function: if $x<y$ in the domain, then $f(x)\le f(y)$. Here, $f\equiv\ln 2$, so for all $x, y\in (0, \infty)$ such that $x\le y$, we have $f(x)=\ln 2\le\ln 2=f(y)$ and hence $f$ is increasing there. Similarly, we obtain $f$ is decreasing there.. – Yuval Gat Apr 20 at 21:03
$$\int_{x}^{2x}{\frac{dt}{t}}=\log{2}$$ therefore it is a constant function or both increasing and decreasing.
Yes, the function is constant and it is always equal to $$\log2$$. So, it is both increasing and decreasing.