Question

The domain of $f\left( x \right) = \dfrac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}}$

Answer 1

Hint : Domains of any function are nothing but the list of all the possible inputs that would make the function value valid. For instance, the value inside logarithm $\log {\rm{ }}x$ can never be less than 0, it means that all the inputs to this logarithmic function should be greater than 0. So, the domain of the function would be $\left( {0,\infty } \right)$ .

Complete step-by-step answer :
The given function is $f\left( x \right) = \dfrac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}}$ .
For a function to be defined, there is a set of input values at which the function can be defined.
We know that a logarithmic function should be greater than 0.
So, $x + 3 > 0$ i.e. $x > - 3......\left( 1 \right)$
Also, in a fractional function, the denominator should be greater than zero for the function to be defined.
So, consider the denominator of the given function and equate it to zero.
${x^2} + 3x + 2 = 0$
Now, we can compare the equation ${x^2} + 3x + 2$ with $a{x^2} + bx + c = 0$ and find the values of $a,b,c$ .
On comparing we get, $a = 1,{\rm{ }}b = 3,{\rm{ }}c = 2$ .
Now, to find the roots of the equation, substitute $a = 1,{\rm{ }}b = 3,{\rm{ }}c = 2$ in $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
After substituting the values, we get,
$\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4 \cdot 1 \cdot 2} }}{{2 \cdot 1}}\\\ = \dfrac{{ - 3 \pm \sqrt {9 - 8} }}{2}\\\ = \dfrac{{ - 3 \pm 1}}{2}$
On simplifying further, we get,
$x = \dfrac{{ - 3 + 1}}{2}\\\ = \dfrac{{ - 2}}{2}\\\ = - 1$
Or
$\Rightarrow x = \dfrac{{ - 3 - 1}}{2}\\\ = \dfrac{{ - 4}}{2}\\\ = - 2$
The roots of the equation ${x^2} + 3x + 2 = 0$ are -1 and 2.
The denominator of the given function becomes zero at $x = - 1$ and $x = 2$ .
The function $f\left( x \right) = \dfrac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}}$ is not defined at $x = - 1$ , $x = 2$ and when $x < \- 3$ .
The function can be defined when $x$ is greater than -3 except for the values $x = - 1$ , $x = -2$ .
So, the domain of the function is \left( { - 3,\infty } \right) - \left\\{ { - 1,2} \right\\}

Therefore, the domain of the given function $f\left( x \right) = \dfrac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}}$ is \left( { - 3,\infty } \right) - \left\\{ { - 1,2} \right\\}.

Note : While finding the domain any function, it is necessary to note that at what point the particular function becomes invalid. For instance, in a fractional function, the function becomes invalid when the denominator becomes zero, such conditions will help students to reach the final solution and avoid confusion in finding the domain of the given function.