I am asked to determine $(f+g)(x)$ and state any restrictions on the domain.
\begin{equation} \begin{split} f(x) & = \sqrt{x-1} \\ g(x) & = -x \end{split} \end{equation}
I have the first part, but I am having trouble determining the restrictions on the domain. \begin{equation} \begin{split} (f+g)(x) & = \sqrt{x-1}+(-x) \\ & = \sqrt{x-1}-x \end{split} \end{equation} My questions are: How could I go about finding the domain for a function like this? Is it as simple as saying as long as the square root is not negative $x\geq0$, $x\in R$ with no other restrictions?
Would the correct
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$\begingroup$In general, if you know the domain of a function $f(x)$ and a function $g(x)$, then the domain of the function $(f + g)(x)$ is $$ \text{(domain of } f(x)) \cap \text{(domain of } g(x)). $$
Here, $f(x) = \sqrt{x - 1}$, and we cannot have any negative numbers under the square root, so we know $x - 1 \geq 0$, which gives $x \geq 1$. In interval notation, this is $[1, \infty)$.
$g(x) = -x$, and we can plug any real number into this, so it is defined everywhere. So the domain of $g(x)$ is $\mathbb{R}$ (all real numbers). You can also write it as $(-\infty, \infty)$.
So the domain of $(f + g)(x)$ is, as we said, the domain of $f(x)$ intersected with the domain of $g(x)$. This is $[1, \infty) \cap (-\infty, \infty)$. If you know how intersections work, then you will know $[1, \infty) \cap (-\infty, \infty) = [1, \infty)$.
Thus, our final answer for the domain of $(f + g)(x)$ is $[1, \infty)$.
$\endgroup$ 3 $\begingroup$$x\geq1$. This is to make sure that the value under the root is non-negative.
$\endgroup$ $\begingroup$You can think of it as:
$$(f+g)(x)=f(x)+g(x)$$
So your domain will be the domain when both $f$ and $g$ are defined simultaneously. In this case you have no problems with $g(x)$ so your domain will be those points where the argument of the root is non negative:
$$x-1 \geq 0 $$
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