find the intervals on which the function is continuous. y = \\sqrt{2x + 2} continuous on the interval (- 1…

Explanation:

Step1: Recall square - root domain

For the function $y = \sqrt{2x + 2}$ to be well - defined in the real number system, the expression inside the square - root must be non - negative. So we set up the inequality $2x+2\geq0$.

Step2: Solve the inequality

Subtract 2 from both sides: $2x\geq - 2$. Then divide both sides by 2, we get $x\geq - 1$. A square - root function $\sqrt{u(x)}$ is continuous for all $x$ in its domain where $u(x)$ is a continuous function. Since $u(x)=2x + 2$ is a linear (and thus continuous) function, the function $y=\sqrt{2x + 2}$ is continuous for all $x$ in its domain.

Answer:

continuous on the interval $[-1,\infty)$