What is the name of the function $f(x)=\frac{1}{x}$?
I'm facing this function:
$$f(x)=\frac{1}{x}$$
What I know is that the above equation is one of the simplest forms of "rational functions", where the numerator is $1$ and the denominator is $x$.
Is its name "harmonic function" too? I have seen it in some papers with unclear explanations!
Comparing $f(x)=1/x$ and $f(x)=(x_o)\exp(-Cx)$ which is an exponential decay function, both give negative rate of change with some shared features, what the main difference between them? advantages and disadvantages?
First one some times diverges to $\infty$, correct?
Thanks
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$\begingroup$You are right: The function defined by $$ f(x) = \frac{1}{x} $$ is a rational function. Some might call this the Harmonic function, but I don't think that this is a good idea because there is a (probably) more common notion of a Harmonic function.
The graph of the function is a Hyperbola.
Now, we do also have something called a Harmonic series. The Harmonic series is the series: $$ \sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots $$
In the end I don't believe that there is a commonly used name for the function $f(x) = \frac{1}{x}$.
You ask what the advantages of the function $f(x) = \frac{1}{x}$ compared to the other function that you list is. It isn't clear what exactly you mean by this. A function is just a function. Only what you know what you want to do can you talk about one function having advantages.
If you for example wanted to model something that decreases, say you want to model the decrease in a population of sorts, then it might turn out that the exponential function "works best" as a model.
$\endgroup$ 1 $\begingroup$The name is $f{}{}{}{}{}{}{}{}{}{}{}{}{}$
$\endgroup$ 1 $\begingroup$This could fairly be called the reciprocal function (the function that sends every real number to its reciprocal), or the multiplicative inverse function. The exponential function $e^{-x}$ has more applications as a physical model for various reasons, including that it remains bounded as $x\rightarrow0$.
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