State the interval(s) over which the function $$ f(x)=\sqrt{x+1} $$ is continuous. Enter your answer in interval notation. To enter $$ \infty $$, type infinity. To enter $$ \cup $$, type $$ U $$.

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To determine the interval(s) over which the function $$ f(x) = \sqrt{x+1} $$ is continuous, we need to consider the domain of the function. The square root function is defined for non-negative values, so we need to find when the expression inside the square root is non-negative:

$$
x + 1 \geq 0
$$

Solving this inequality:

$$
x \geq -1
$$

This means that the function $$ f(x) $$ is defined and continuous for all $$ x $$ greater than or equal to $$-1$$. Therefore, the interval of continuity is:

$$
[-1, \infty)
$$

In interval notation, the answer is:

$$
[-1, infinity)
$$
The function $$ f(x) = \sqrt{x+1} $$ is continuous on the interval $$[-1, \infty)$$.

State the interval(s) over which the function \( f(x)=\sqrt{x+1} \) is continuous. Enter your answer in interval notation. To enter \( \infty \), type infinity. To enter \( \cup \), type \( U \).

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