Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

up vote 0 down vote accepted

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.

share|improve this answer

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 $(f+g)(x)$.

enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.