Question

How do you find the domain and range of $f(x) = \dfrac{{{x^2} + 5x - 6}}{{{x^2} - 5x + 6}}$ ?

Answer 1

The domain of a function is the complete step of possible values of the independent variable. That is 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. The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain.

Complete step by step solution:
Given, $f(x) = \dfrac{{{x^2} + 5x - 6}}{{{x^2} - 5x + 6}}$. The given function in the form of fraction, we can never divide by zero. To find the domain by setting the denominator $\left( {{x^2} - 5x + 6} \right)$ equal to zero and solving for x, you can calculate the values that will be excluded in the function. Let consider the denominator which equate to zero
${x^2} - 5x + 6$
This can be solve for x by using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where $b = - 5$, $a = 1$ and $c = 6$, then
$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( 6 \right)} }}{{2\left( 1 \right)}}$
$\Rightarrow \,\,\,\,x = \dfrac{{5 \pm \sqrt {25 - 24} }}{2}$
$\Rightarrow \,\,\,\,x = \dfrac{{5 \pm \sqrt 1 }}{2}$
$\Rightarrow \,\,\,\,x = \dfrac{{5 \pm 1}}{2}$
$\Rightarrow \,\,\,\,x = \dfrac{{5 + 1}}{2}$ or $x = \dfrac{{5 - 1}}{2}$
On simplification, we get
$x = 3$ or $x = 2$
This means that, when $x = 2$ or 3, we have the denominator $\,{x^2} - 5x + 6 = 0$. It implies that $f\left( x \right) = \dfrac{x}{0}$ which is undefined.Hence, the domain is all real numbers except 2 and 3. Also, the domain of given function can be written as:
${D_f} = \left( { - \infty ,2} \right) \cup \left( {2,3} \right) \cup \left( {3, + \infty } \right)$
Or
{D_f} = \mathbb{R} - \left\\{ {2,\infty } \right\\}

To Find the Range, Say the given function $f\left( x \right) = y$ and rearrange the function as a quadratic equation.
$y = \dfrac{{{x^2} + 5x - 6}}{{{x^2} - 5x + 6}}$
$\Rightarrow \,\,y\left( {{x^2} - 5x + 6} \right) = {x^2} + 5x - 6$
Take RHS to the LHS, then
$y{x^2} - {x^2} - 5yx - 5x + 6y + 6 = 0$
$\Rightarrow \,\,\left( {y - 1} \right){x^2} - 5\left( {y + 1} \right)x + 6\left( {y + 1} \right) = 0$
We know the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, that the solutions of $x$ are real when ${b^2} - 4ac \geqslant 0$
So likewise, we can written as $25{\left( {y + 1} \right)^2} - 24\left( {y - 1} \right)\left( {y + 1} \right) \geqslant 0$
Now, we solve the inequality for the values set of values of $y$
$25\left( {{y^2} + 2y + 1} \right) - 24\left( {{y^2} - 1} \right) \geqslant 0$
$\Rightarrow \,\,25{y^2} + 50y + 25 - 24{y^2} + 24 \geqslant 0$
On simplification, we get
${y^2} + 50y + 49 \geqslant 0$
$\Rightarrow \,\,{y^2} + y + 49y + 49 \geqslant 0$
$\Rightarrow \,\,y\left( {y + 1} \right) + 49\left( {y + 1} \right) \geqslant 0$
$\Rightarrow \,\,\left( {y + 1} \right)\left( {y + 49} \right) \geqslant 0$
$\Rightarrow \,\,\left( {y + 1} \right) \geqslant 0$ or $\left( {y + 49} \right) \geqslant 0$
$\therefore\,\,y \geqslant - 1$ or $y \geqslant - 49$

Hence, the range is $\left( { - \infty , - 49} \right] \cup \left[ { - 1, + \infty } \right)$ and the domain is all real numbers except 2 and 3.

Note: The domain where the x values ranges and the range where the y values ranges. Since the equation is a quadratic equation we find the value of x and then we determine the value of y for the values of x. We must the know the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to determine the value of x and it is applicable only to this problem since it contains quadratic equation.