What is the domain of the function $f(x)=\sqrt{\frac{1}{3} x+2}$?
To find the domain of the function \(f(x) = \sqrt{\frac{1}{3}x + 2}\), we must determine the set of all real values of \(x\) for which the function is defined. For a square root function \(\sqrt{u}\) to yield a real value, the expression under the square root must be non-negative, meaning \(u \geq 0\). In this specific case, our expression under the root is \(u = \frac{1}{3}x + 2\).
We set the expression inside the square root to be greater than or equal to zero:
\[\frac{1}{3}x + 2 \geq 0\]
To solve for \(x\), we first isolate the term containing \(x\) by subtracting 2 from both sides of the inequality:
\[\frac{1}{3}x \geq -2\]
Now, we multiply both sides of the inequality by 3. Since 3 is a positive number, the direction of the inequality remains the same:
\[x \geq -6\]
This result means the function \(f(x)\) is defined for all values of \(x\) that are greater than or equal to -6.
Comparing our result \(x \geq -6\) to the given options:
The calculated domain matches option d. Therefore, the domain of the function is \(x \geq -6\).
The domain of a function represents the range of possible input values for a specific model. For instance, if you are designing a structural beam where the strength depends on its length \(x\) through a square root model like \(f(x) = \sqrt{\frac{1}{3}x + 2}\), you need to know the minimum length \(x\) for which the beam is physically stable. By determining the domain, you identify that the beam must be at least 6 units long in the negative direction (or relative to a specific reference point) for the model to remain valid, helping engineers build safe structures.