Question

How do you find the domain of $f\left( x \right) = \dfrac{{8x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}$ ?

Answer 1

Here in this question, we need to find the domain of the given function. The domain of a function is the complete step of possible values of the independent variable. For this, the denominator of a given function $f\left( x \right)$ cannot be zero as this would make $f\left( x \right)$ undefined then equating the denominator to zero and solving gives the values that x cannot be zero.

Complete step by step answer:
The domain is the set of all possible ‘x’ values which will make the function ‘work’ and will give the output of ‘y’ as a real number.Given,
$f\left( x \right) = \dfrac{{8x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}$
The given function in the form of fraction, we can never divide by zero.
To find the domain by setting the denominator $\left( {x - 1} \right)\left( {x - 2} \right)$ equal to zero and solving for x, we can calculate the values that will be excluded in the function.

Let consider the denominator which equate to zero
$\Rightarrow \,\,\,\left( {x - 1} \right) = 0$ or $\left( {x - 2} \right) = 0$
On simplification, we get
$\Rightarrow \,\,\,\,x = 1$ or $x = 2$
This means that, when $x = 1$ or 3, we have the denominator $\left( {x - 1} \right)\left( {x - 2} \right) = 0$
It implies that $f\left( x \right) = \dfrac{x}{0}$ which is undefined.
Hence, the domain is all real numbers except 1 and 2.
Also, the domain of given function can be written as:
${D_f} = \left( { - \infty ,1} \right) \cup \left( {3, + \infty } \right)$

Note: The domain where the x values range and the range where the y values ranges.If the function there are no radicals or fractions involved. In this case, there is no real number that makes the expression undefined. If the function is in fraction, set the denominator equal to zero and exclude the x value you find when you solve the equation.