Question

Find the domain of the function $f\left( x \right)=\dfrac{1}{{{\log }_{10}}\left( 1-x \right)}+\sqrt{x+2}$ .

Answer 1

Hint : We will first consider $f\left( x \right)$ as $g\left( x \right)+h\left( x \right)$ . Then we will take $g\left( x \right)$ as $\dfrac{1}{{{\log }_{10}}\left( 1-x \right)}$ and $h\left( x \right)$ as $\sqrt{x+2}$ . After that we will find the domains of both and finally find the common values of the domain, which is the domain of $f\left( x \right)$ .

Complete step-by-step answer :
In the question, we are given a function, $f\left( x \right)$ , which is $\dfrac{1}{{{\log }_{10}}\left( 1-x \right)}+\sqrt{x+2}$ and we have been asked to find the domain of the function. Before we start solving this question, let us find out what a domain is. In mathematics, the domain or set of departure of a function is the set to which all of the input is constrained to fall. It is the set X in the notation, $f:X\to Y$ . Since, a function is defined on its entire domain, it coincides with its definition. However, this coincidence is no longer true for a partial function and can be a proper subset of the domain. A domain is a part of a function $f$ , if it is defined as a triple ( X, Y, G), where X is called the domain of $f$ , Y its co-domain and G its graph. A domain is not part of a function if $f$ it is defined just as a graph. For example, in the set theory, it is desirable to permit the domain of a function to be a proper class X, in which case there is no such thing as a triple (X, Y, G). With such a definition functions do not have a domain, although some authors still use it informally after introducing a function in the form $f:X\to Y$ . For instance, the domain of cosine is the set of all real numbers while the domain of the square root consists only of the numbers greater than or equal to 0.
Now, here, let us say $f\left( x \right)=g\left( x \right)+h\left( x \right)$ , where $g\left( x \right)$ is $\dfrac{1}{{{\log }_{10}}\left( 1-x \right)}$ and $h\left( x \right)$ is $\sqrt{x+2}$ . So, then we will find the domain of the functions separately and find the common values of x. So, for,
$g\left( x \right)=\dfrac{1}{{{\log }_{10}}\left( 1-x \right)}$
Here, $1-x>0\text{ }and\text{ }{{\log }_{10}}\left( 1-x \right)\ne 0$ .
So, $\begin{aligned} & x<1\text{ }and\text{ }\left( 1-x \right)\ne 10{}^\circ \\\ & or,\text{ }x<1\text{ }and\text{ }\left( 1-x \right)\ne 1 \\\ & or,\text{ }x<1\text{ }and\text{ }x\ne 0 \\\ \end{aligned}$
Hence, the domain will be, \left( -\infty ,1 \right)-\left\\{ 0 \right\\} . Now, to find the other domain,
$h\left( x \right)=\sqrt{x+2}$
Here, $x+2\ge 0$ or $x\ge -2$ .
Hence, the domain will be $[-2,\infty )$ . We can see that the common values of x or the domain of $f\left( x \right)$ will be $[-2,1)$ .
Therefore, the domain is [-2,1)-\left\\{ 0 \right\\} .

Note : The domain in simple words can also be said as those values for which a defined value of $f\left( x \right)$ exists. We can also write the domain, [-2,1)-\left\\{ 0 \right\\} as $[-2,0)\cup \left( 0,1 \right)$ .