How do you find the domain and range algebraically of f(x)=1/sqrt(x-4)f(x)=1x4?

1 Answer
Feb 7, 2015

Your function has a square root but also this square root is in the denominator of a fraction.

So, what you wan is to avoid values of xx that:

1) make the argument of the square root NEGATIVE (you cannot find a real number as solution of a negative square root);
2) make the denominator equal to ZERO (you cannot evaluate a division by zero).

To express the above points mathematically you write that x-4x4 MUST be >0>0.

This means that you can accept only values that satisfy:
x-4>0x4>0
x>4x>4

44 is not acceptable because if you use it you get sqrt(4-4)44 that can be calculated and gives 00 but it would be in the denominator and this is not acceptable.
When you get near 44 your function gets very big tending to +oo+ while going towards xx values very big the function tends to zero (you can try by yourself substituting values in your function and calculating it).
So your domain will be all the positive xx bigger than 44 up to +oo+.
With a range for y=f(x)y=f(x) of +oo+ to 00.
Or graphically:

graph{1/sqrt(x-4) [-10, 10, -5, 5]}