Question

How do you find the domain of $ y = \sqrt {{x^2} - 6x + 5} $ ?

Answer 1

Hint: The value under the root should always be greater or equal to zero for a positive and real outcome, so we can say that the value of $ \sqrt {{x^2} - 6x + 5} $ will be always positive or equal to zero and for $ \sqrt {{x^2} - 6x + 5} $ to be positive the value of $ {x^2} - 6x + 5 $ should be equal to or greater than $ 0 $ , so the domain can be found as,
Domain, Df:
 $ {x^2} - 6x + 5 \geqslant 0 $

Complete step-by-step answer:
In order to find the domain of the given equation we first have to find the zeroes as the value of y changes its sign at zeroes of the equation, so for zeroes we will equate our equation to zero.
 $ {x^2} - 6x + 5 = 0 $
To calculate zeroes, we have the quadratic formula as,
 $ {x_{1,2}} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Comparing our equation with the standard quadratic equation, $ a{x^2} + bx + c = 0 $ we will have,
 $ a = 1 $ , $ b = - 6 $ and $ c = 5 $
Putting these values in our quadratic formula we will get,
 $ {x_{1,2}} = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(1)(5)} }}{{2(1)}} $
Simplifying the values inside root will give us,
 $ {x_{1,2}} = \dfrac{{ - ( - 6) \pm \sqrt {36 - 20} }}{{2(1)}} $
As $ 36 - 20 = 16 $ and $ \sqrt {16} = 4 $ putting this in above equation,
 $ {x_{1,2}} = \dfrac{{ - ( - 6) \pm 4}}{{2(1)}} $
Also, $ - ( - 6) = 6 $
 $ {x_{1,2}} = \dfrac{{6 \pm 4}}{2} $
Now the two roots will be, \[{x_1} = \dfrac{{6 + 4}}{2}\]and $ {x_2} = \dfrac{{6 - 4}}{2} $ , solving them further will give us,
\[{x_1} = 5\] and $ {x_2} = 1 $
So at these two roots, the sign of the equation will change.
So now the whole X-axis is divided in three parts $ [ - \infty ,1] $ , $ [1,5] $ and $ [5,\infty ] $ . As per our discussion we have to find the positive values of the function, in order to find that we can choose any one point in each part and can evaluate the value of the function.
Let us find the value of function at $ 0 $ ,
 $ f(0) = {x^2} - 6x + 5 = 5 $ , which is positive
So,if $ x \in [ - \infty ,1] $ will yield positive results, and as $ 1 $ is the root so sign will get changed here and $ x \in [1,5] $ will give us negative and again $ x \in [5,\infty ] $ will give us positive values.
As there is root on the function we will only consider positive values so our required answer will be,
 $ x \in [ - \infty ,1] $ and $ x \in [5,\infty ] $ only.

Note: If our function yields a negative value inside the root then it will give imaginary values which are undesirable so we have to only consider the positive values of the function.