## Precalculus: Mathematics for Calculus, 7th Edition

$\{x\in R: x\leq \frac{7}{3}\}=[-\infty , \frac{7}{3}]$
The domain of $g(x)=\sqrt {7-3x}$ is the set of all real numbers $x$ for which $\sqrt{7-3x}$ is defined as a real number. We know the sqare root of a negative number is not defined, so the domain of $g(x)=\sqrt{7-3x}$ is all real numbers $x$ for which $7-3x \geq 0$. So we need to solve this inequality for $x$. We add $3x$ to both sides to get $$7 \geq 3x,$$ and dividing both sides gives $$\frac{7}{3} \geq x,$$ or equivalently, $$x \leq \frac{7}{3}$$Thus the domain of this function is all real numbers $x$ for which $x\leq \frac{7}{3}$ In set notation, the domain is $\{x\in R: x\leq \frac{7}{3}\}.$ In interval notation, the domain is $[-\infty , \frac{7}{3}].$ - there is a bracket after $\frac{7}{3}$ because $\frac{7}{3}$ IS included in the equation whereas a parentheses means that the number is NOT included in the question.