Question

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

Answer 1

`DinR-[1,3]`

Explanation:

`f(x)=1/sqrt(x^2-4x+3)`

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

i.e
`f(x)` is discontinuous `rarr f(x)=1/0`

`1/sqrt(x^2-4x+3)=1/0`

`sqrt(x^2-4x+3)=0`

By squaring both sides

`x^2-4x+3=0`

`(x-3)(x-1)=0`

`x=1 " , "x=3`

Also the term

`x^2-4x+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[`

`color(green)("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 `[1,3]`

`DinR-[1,3]`

`color(blue)(S"econd"`

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

`color(blue)("Or")`

• simply by looking at the function

`y=1/sqrt(x^2-4x+3)`

the denominator can neither be negative nor zero

because the term `sqrt(x^2-4x+3)>0`

and so

`y>0`

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