at what points is the following function continuous?\n\n$f(x)=\\sqrt{3x + 5}$\n\nthe set of x - values where…

Explanation:

Step1: Determine the domain of the function

For the square - root function (y = \sqrt{u}), the expression inside the square root (u) must be non - negative. Here, (u=3x + 5). So, we set up the inequality (3x+5\geq0). [ \begin{align*} 3x+5&\geq0\ 3x&\geq - 5\ x&\geq-\frac{5}{3} \end{align*} ]

Step2: Recall the continuity of square - root functions

The square - root function (y = \sqrt{u}) is continuous for all (u\geq0). Since (u = 3x + 5) is a linear function (and linear functions (y=ax + b) are continuous for all real (x)), and the composition of a continuous function (y=\sqrt{u}) (continuous for (u\geq0)) with a continuous function (u = 3x+5) (continuous for all (x\in R)) is continuous for all (x) in the domain of the composition.

Answer:

(\left[-\frac{5}{3},\infty\right))