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

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 $x$ 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 $x - 4$ MUST be $> 0$.

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

$4$ is not acceptable because if you use it you get $\sqrt{4 - 4}$ that can be calculated and gives $0$ but it would be in the denominator and this is not acceptable.
When you get near $4$ your function gets very big tending to $+ \infty$ while going towards $x$ 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 $x$ bigger than $4$ up to $+ \infty$.
With a range for $y = f \left(x\right)$ of $+ \infty$ to $0$.
Or graphically:

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