Question
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 \( U \).
Ask by Davey Mitchell. in the United States
Mar 12,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The function \( f(x) = \sqrt{x+1} \) is continuous on the interval \([-1, \infty)\).
Solution
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:
\[
[-1, infinity)
\]
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Simplify this solution Mind Expander
The function \( f(x) = \sqrt{x+1} \) is continuous for all values of \( x \) where the expression under the square root is non-negative. This means \( x+1 \geq 0 \), which simplifies to \( x \geq -1 \). Therefore, the function is continuous on the interval \([-1, \infty)\).
