Find the domain of the function f(x) = sqrt(x + 10). Give your answer in interval notation.

Step-by-step solution

1. 1

   Identify the single restriction. A square root of a negative number is not a real number, so the only requirement is that the expression under the radical — the radicand — be nonnegative. There is no denominator here, so unlike $1/\sqrt{x+10}$ a radicand of exactly zero is allowed, and that distinction decides whether the endpoint is included.

2. 2

   Write the inequality.

   $x+10\geq 0$

3. 3

   Solve it. Subtract $10$ from both sides; since this is addition rather than multiplication by a negative, the inequality direction is unchanged:

   $x\geq -10$

4. 4

   Express the answer in interval notation. The set of all $x$ at least $-10$ is

   $[-10,\infty)$

   The square bracket marks $-10$ as included, and the parenthesis at $\infty$ reflects that infinity is not a number to be attained.

5. 5

   Check the endpoint and one value on each side. At $x=-10$, $f(-10)=\sqrt{0}=0$, a perfectly valid real output, so the closed bracket is right. At $x=-9$, $f(-9)=\sqrt{1}=1$. At $x=-11$ the radicand is $-1$ and $f$ is undefined over the reals, confirming that nothing to the left of $-10$ belongs to the domain.