# Find the domain f(x) =1 /sqrtx^2-4x+3?

May 4, 2018

$D \in R - \left[1 , 3\right]$

#### Explanation:

$f \left(x\right) = \frac{1}{\sqrt{{x}^{2} - 4 x + 3}}$

The function is discontinuous at certain points which you can get by equalling the denominator of the function to Zero

i.e
$f \left(x\right)$ is discontinuous $\rightarrow f \left(x\right) = \frac{1}{0}$

$\frac{1}{\sqrt{{x}^{2} - 4 x + 3}} = \frac{1}{0}$

$\sqrt{{x}^{2} - 4 x + 3} = 0$

By squaring both sides

${x}^{2} - 4 x + 3 = 0$

$\left(x - 3\right) \left(x - 1\right) = 0$

$x = 1 \text{ , } x = 3$

Also the term

${x}^{2} - 4 x + 3$ can't be negative due to the root

and You can find where the function is negative using its graph

graph{x^2-4x+3 [-0.654, 5.508, -1.507, 1.57]}

And So the function doesn't exist from ]1,3[

$\textcolor{g r e e n}{\text{you could also try to substitute with values in the function to find which part is positive and which is negative}}$

color(blue)(x<1 " and " x>3" will give a positive value"

color(red)("1< "x<3 " will give negative value"

And at last with the function's graph

graph{1/sqrt(x^2-4x+3) [-2.603, 7.27, -1.28, 3.65]}

from the graph the function doesn't exist on $\left[1 , 3\right]$

$D \in R - \left[1 , 3\right]$

color(blue)(S"econd"

• Its Range can be found through the graph and it will be ]0,oo[

$\textcolor{b l u e}{\text{Or}}$

• simply by looking at the function

$y = \frac{1}{\sqrt{{x}^{2} - 4 x + 3}}$

the denominator can neither be negative nor zero

because the term $\sqrt{{x}^{2} - 4 x + 3} > 0$

and so

$y > 0$

color(green)("as 1 divided by any positive number will give a positive value"