Find the Domain and Range of \(f(x)=\sqrt{x-3}\)
Question:
Find the domain and range of the real valued function:
$$
f(x)=\sqrt{x-3}
$$
Solution
Domain
For a square root function, the expression inside the root must be non-negative.
Therefore, $$ x-3\ge0 $$
$$ x\ge3 $$
Hence, the domain is: $$ [3,\infty) $$
Range
Since square root values are always non-negative, $$ f(x)\ge0 $$
Minimum value occurs at $$ x=3 $$
$$ f(3)=\sqrt{3-3}=0 $$
As \(x\) increases, \(f(x)\) increases without bound.
Hence, the range is: $$ [0,\infty) $$