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

#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"#