Find the Domain and Range of \(f(x)=\sqrt{x-1}\)
Question:
Find the domain and range of the real valued function:
$$
f(x)=\sqrt{x-1}
$$
Solution
Domain
For a square root function, the expression inside the root must be non-negative.
Therefore, $$ x-1\ge0 $$
$$ x\ge1 $$
Hence, the domain is: $$ [1,\infty) $$
Range
Since square root values are always non-negative, $$ f(x)\ge0 $$
Minimum value occurs at $$ x=1 $$
$$ f(1)=\sqrt{1-1}=0 $$
As \(x\) increases, \(f(x)\) increases without bound.
Hence, the range is: $$ [0,\infty) $$