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

##### 1 Answer

#### Explanation:

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

i.e

**discontinuous**

**By squaring both sides**

**Also the term**

**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**

**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**

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

- simply by looking at the function

the denominator can neither be negative nor zero

because the term

and so