# Finding domains?

I need to find what $(f + g)(x)$ would equal here..

$$f(x) = \sqrt{25 − x^2},\quad g(x) = \sqrt{x^2 − 4}$$

Am I supposed to just add $\sqrt{25 − x^2}$ and $\sqrt{x^2 − 4}$? And if so, how? And how would I then determine the domain?

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The defintion for $f+g$ is pointwise, that is, $(f+g)(x)=f(x)+g(x)$, for all $x$ in the intersection of the domains. – Sigur Sep 25 '12 at 23:59

If $f(x)=\sqrt{25-x^2}$ and $g(x)=\sqrt{x^2-4}$, then $(f+g)(x)=\sqrt{25-x^2}+\sqrt{x^2-4}$, so yes: to find the value of $f+g$ at any point $x$, just add the values of $f(x)$ and $g(x)$.

In the context of this question the domain of $f+g$ is the set of all real numbers $x$ for which $(f+g)(x)$ can be calculated as a real number. In other words, you need to be able to take both of those square roots. This means that you need to have $25-x^2\ge 0$ and $x^2-4\ge 0$: you have to find those values of $x$ that satisfy both inequalities.

There are several ways to do this, but I’d begin by rewriting them as $x^2\le 25$ and $x^2\ge 4$. $x^2\le 25$ if and only if $|x|\le 5$, which in turn is true if and only if $-5\le x\le 5$. Similarly, $x^2\ge 4$ if and only if $|x|\ge 2$, which is true if and only if $x\le-2$ or $x\ge 2$. Thus, you need to find all those real numbers $x$ such that

$$-5\le x\le 5\quad\mathbf{and}\qquad\Big(x\le -2\quad\mathbf{or}\quad x\ge 2\Big)\;.\tag{1}$$

At this point it’s just a matter of converting $(1)$ into a simpler, more readily understandable form.

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Here is the graphs. I used GeoGebra to plot. The red one is for $f(x)=\sqrt{25-x^2}$; the blue one is for $g(x)=\sqrt{x^2-4}$; the green one is for the sum $(f+g)(x)$.

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