Calculate the domain of Exponential function with variable base

Question

I'm trying to find the domain of the following function:

$$f(x)={\Biggl(\frac{\displaystyle\ {\log}^{2}(2\cos(x)-1) + e^{x}}{\displaystyle \sqrt{1-\sin^2(x)} - \sin(x)}\Biggr)}^{\arcsin\sqrt{x}}$$

I have reasoned this way: since I have an exponential function with variable base, this last one must be posed $>0$. And so:

$$\frac{\displaystyle\ {\log}^{2}(2\cos(x)-1) + e^{x}}{\displaystyle \sqrt{1-\sin^2(x)} - \sin(x)}>0$$

Even the exponent has an its condition which is: $$0\le\sqrt{x}\le1$$

Hence I have to solve the following system:

$$ \left\{ \begin{array}{c} \frac{\displaystyle\ {\log}^{2}(2\cos(x)-1) + e^{x}}{\displaystyle \sqrt{1-\sin^2(x)} - \sin(x)}>0 \\ 0\le\sqrt{x}\le1 \end{array} \right. $$

This is the point in which I get blocked, I do not know how to solve the first inequality of the system: specifically, I'm having troubles in solving the numerator: $${\log}^{2}(2\cos(x)-1) + e^{x}>0$$ I tried to solve it by using graphic mode but I found it not so accurate.

Now, someone could say me if my calculus is wrong or could give me another way (or hint) to determine the domain of $f(x)$? Thank you.

Answer 1

Numerator is not a problem, since it is the sum of two positive quantities $${\log}^{2}(2\cos x -1) + e^{x}>0, \forall x\in\mathbb{R}$$

except the values which give negative or null $2\cos x -1$

Notice that $\sqrt{1-\sin^2 x} - \sin(x)=\cos x -\sin x$

Argument of logarithm, denominator and exponent lead to the system

$ \left\{ \begin{array}{l} 2\cos x -1>0\\ \cos x - \sin x >0 \\ 0\le\sqrt{x}\le 1\\ \end{array} \right. $

$ \left\{ \begin{array}{l} 0\le x \le 1\\ \cos x >\dfrac{1}{2}\\ \dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x} >0 \end{array} \right. $

$ \left\{ \begin{array}{l} 0\le x \le 1\\ \cos x >\dfrac{1}{2}\to -\dfrac{\pi}{3}+2k\pi< x <\dfrac{\pi}{3}+2k\pi,\,k\in\mathbb{Z}\\ \tan x < 1 \to h\pi\le x <\dfrac{\pi}{4}+h\pi,\,h\in\mathbb{Z} \end{array} \right. $

which gives the conditions

$\color{red}{0\leq x<\dfrac{\pi}{4}}$

Answer 2

Strictly speaking, domain is to be provided along with the function, i.e. it is part of the function definition. e.g. $f_1(x) = x^2, x\in \mathbb{Z}$, and $f_2(x) = x^2, x\in \mathbb{R}$ are two different functions.

Having said that, "finding domain" is a high-school level exercise where it is convention is to find the largest subset of $\mathbb{R}$ for which an algebraic function is well defined.

"Exponentials" pose the biggest challenge in this exercise due to ambiguous definitions or missing conventions. We will hence first need to set the ground rules before we may talk about answering the question.

"Exponential Functions" (EF for short) is a term, reasonably well defined, where the base is positive constant, i.e. this is a set of functions of the form $a^x, a>0, x\in\mathbb{R}$ where $a$ is a constant. Here the domain is $\mathbb{R}$.

Unfortunately, this definition casts its shadow on wider set of functions of the type ${f(x)}^{g(x)}$ for the purpose of domain-finding-exercise. I wouldn't dare call these functions as "Exponential Functions", but lets give them a name, say, "Variable Exponent Functions" or VEF.

One of the simplest of VEF would be $f_3(x) = (-1)^x$. What is its domain? I have seen some purist say that since this is an EF violating EF rule of $a>0$ so domain is $\phi$. Then there are people in the "maximal-domain-camp" saying that domain is $\{x\in\mathbb{Q},x=p/q,p\in\mathbb{Z},q\in\mathbb{Z_{odd}}\}$, where $\mathbb{Z_{odd}}$ is the set of all odd integers. I have found Mathematicians not really interested in setting ground rules for high-school exercises, but we will need some to move forward. I would go further with the "maximal-domain-camp" as this keeps the essence of domain-finding exercise alive.

Before we move further, there is still an issue of definition of operator precedence in exponents. I have not seen a convention agreed by all. One detailed discussion is here: What are the Laws of Rational Exponents?. Discussion goes into validity of cases like $x^{6/15}=x^{2/5}$ and $x^{6/6}=x$ and different definitions of exponentiations in algebra and calculus. Again I am not sure if people agree about those definitions within those domains either.

So the answer to the question would depend on which convention you wish to follow (or your teacher follows or thinks is correct).