Find the domain of the following function.
$g(x)=\frac{3}{\sqrt{9+x}}$
The domain is □ .
To find the domain of the function \(g(x) = \frac{3}{\sqrt{9+x}}\), we must identify all values of \(x\) for which the function is well-defined. This requires satisfying two primary mathematical constraints. First, the expression inside the square root, \(9 + x\), must be non-negative for the square root to be a real number. Thus, we require \(9 + x \ge 0\). Second, since the square root is in the denominator, the entire expression \(\sqrt{9 + x}\) cannot equal zero. Therefore, \(9 + x eq 0\), which implies \(\sqrt{9 + x} > 0\).
By combining the two constraints, we see that the expression \(9 + x\) must be strictly greater than zero to ensure the square root is defined and the denominator is not zero. We set up the inequality:
\[9 + x > 0\]
To solve for \(x\), we subtract \(9\) from both sides:
\[x > -9\]
This indicates that the function is defined for all real numbers \(x\) greater than \(-9\). In interval notation, this domain is represented as \((-9, \infty)\).
The domain of the function \(g(x) = \frac{3}{\sqrt{9+x}}\) is the set of all real values of \(x\) such that \(x > -9\). Thus, the domain is \(\boxed{(-9, \infty)}\).
In engineering, such domain constraints are essential when modeling physical systems. For example, if \(x\) represents the time elapsed since an experiment began, and the function \(g(x)\) models the concentration of a chemical, a domain of \(x > -9\) might represent that the sensor began recording 9 seconds before the main reaction initiated at \(t=0\). Ensuring the mathematical function is defined only within relevant physical boundaries helps avoid errors in simulations and data analysis.