Question

Find the domain of this function and how would you plot points of f: f(x) = x/(x^2 + a)

Answer 1

The domain of a function is defined as the set of inputs in which this function is valid. For functions an element in the domain cannot be matched with two different elements in the range (or set of outputs). So we have the following functions:

$f(x)=\frac{x}{x^{2}+a}$

So we need to find the valid inputs. In the numerator we have the function $x$ that is valid for the whole set of the real number. The denominator has the function $x^{2}+a$, so given that this function is in the denominator then it can not be equal zero, therefore:

$If \ a \geq 0 \\ x^{2}+a \neq 0 \ Always! \\ \\ If \ a\ \textless \ 0 \\ x^{2}+a = 0 \ when \ x= \pm \sqrt{a}$

So the domain we will have two options for the domain of the function, namely:

First option:

$If \ a \geq 0 \\ D:\{x \in \mathbb{R}\}$

Second option:

$If \ a\ \textless \ 0 \\ D:\{x \in \mathbb{R}\backslash \ x \neq \pm \sqrt{a}\}$

How to plot points?

For the first option, let $a$ be a real value where $a\geq 0$ and let $x$ take some real values and find the value of $f(x)$. For example:

$Let \ a=2 \\ \\ f(x)= \frac{x}{x^2+2} \\ \\ x=-3, \ f(3)=-0.27 \\ P_{1}(-3,-0.27) \\ \\ x=-2, \ f(-2)=-0.33 \\ P_{2}(-2,-0.33) \\ \\ x=-1, \ f(-1)=-0.33 \\ P_{3}(-1,-0.33) \\ \\ x=0, \ f(0)=0 \\ P_{4}(0,0) \\ \\ x=1, \ f(1)=0.33 \\ P_{5}(1,0.33) \\ \\ x=2, \ f(2)=0.33 \\ P_{6}(2,0.33) \\ \\ x=3, \ f(3)=0.27 \\ P_{7}(3,0.27)$

The points of this function are shown in the figure below. You can do the same for the second option, the only exception is that $x \neq \pm \sqrt{a}$, that is, the only points you must not graph is when $x=\pm \sqrt{a}$. Those points does not exist.