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Find the domain of $f(x)=\dfrac{3x+1}{\sqrt{x^2+x-2}}$

This is my work so far: $$\dfrac{3x+1}{\sqrt{x^2+x-2}}\cdot \sqrt{\dfrac{x^2+x-2}{x^2+x-2}}$$ $$\dfrac{(3x+1)(\sqrt{x^2+x-2})}{x^2+x-2}$$ $(3x+1)(\sqrt{x^2+x-2})$ = $\alpha$ (Just because it's too much to type) $$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{1-4(1)(-2)}}{2}\right]}$$ $$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{9}}{2}\right]}$$ $$\dfrac{\alpha}{\left[\left(\dfrac{-1+3}{2}\right)\left(\dfrac{-1-3}{2}\right)\right]}$$ $$\dfrac{\alpha}{(1)(-2)}$$ Now, I checked on WolframAlpha and the domain is $x\in \mathbb R: x\lt -2$ or $x\gt 1$ But my question is, what do I do with the top of the problem? Or does it just not matter at all.

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How do the $x$'s disappear from the bottom of your fraction? – Code-Guru Aug 7 '12 at 0:25
I input my quadratic formula into the quadratic equation to find the two values of $x$. – Austin Broussard Aug 7 '12 at 0:26
Using the quadratic formula is exactly what you need to do. However, putting the solutions on the bottom of your fraction is incorrect. Also, this doesn't answer the original question about the domain. – Code-Guru Aug 7 '12 at 0:29
The domain is $x\lt -2$ and $x\gt 1$ – Austin Broussard Aug 7 '12 at 0:29
Okay, how do you use the solutions of the quadratic equation to determine that? Hint: why are you using the quadratic equation to begin with? – Code-Guru Aug 7 '12 at 0:30
up vote 6 down vote accepted

Note that $x^2+x-2=(x-1)(x+2)$. There is a problem only if $(x-1)(x+2)$ is $0$ or negative. (If it is $0$, we have a division by $0$ issue, and if it is negative we have a square root of a negative issue.)

Can you find where $(x-1)(x+2)$ is $0$? Can you find where it is negative? Together, these are the numbers which are not in the domain of $f(x)$.

Or, to view things more positively, the function $f(x)$ is defined precisely for all $x$ such that $(x-1)(x+2)$ is positive.

Remark: Let $g(x)=x^2+x-2$. We want to know where $g(x)$ is positive. By factoring, or by using the Quadratic Formula, we can see that $g(x)=0$ at $x=-2$ and at $x=1$.

It is a useful fact that a nice continuous function can only change sign by going throgh $0$. This means that in the interval $(-\infty, -2)$, $g(x)$ has constant sign. It also has constant sign in $(-2,1)$, and also in $(1,\infty)$.

We still don't know which signs. But this can be determined by finding $g(x)$ at a test point in each interval. For example, let $x=-100$. Clearly $g(-100)$ is positive, so $g(x)$ is positive for all $x$ in the interval $(-\infty,0)$.

For the interval $(-2,1)$, $x=0$ is a convenient test point. Note that $g(0) \lt 0$, so $g(x)$ is negative in the whole interval $(-2,1)$. A similar calculation will settle things for the remaining interval $(1,\infty)$.

There are many other ways to handle the problem. For example, you know that the parabola $y=(x+2)(x-1)$ is upward facing, and crosses the $x$-axis at $x=-2$ and $x=1$. So it is below the $x$-axis (negative) only when $-2 \lt x \lt 1$.

Or we can work with pure inequalities. The product $(x+2)(x-1)$ is positive when $x+2$ and $x-1$ are both positive. This happens when $x \gt 1$. The product is also positive when $x+2$ and $x-1$ are both negative. This happens when $x \lt -2$.

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So is all of my work correct? – Austin Broussard Aug 7 '12 at 0:21
@AustinBroussard: The beginning, where you rationalize the denominator, is correct, but unnecessary. In fact it introduces a complication, because now the numerator and denominator are both $0$ at $x=1$ and $x=-2$. For the rest, it is not clear what you are doing. Looks like you are finding the roots of the polynomial at the bottom. the roots are correct, but the expression $\frac{\alpha}{\text{stuff}}$ is not. – André Nicolas Aug 7 '12 at 0:29
Okay, so where do I start then? – Austin Broussard Aug 7 '12 at 0:30
On the graph, there are no $x$ values from $-2$ to $1$. So the $x$ values continue infinitely in either direction from $-2$ and $1$. Right? – Austin Broussard Aug 7 '12 at 0:33
I have added my answer which hopefully adds to @Andre's. I hope it gets you headed in the right direction. Please come back with more questions if you get stuck. – Code-Guru Aug 7 '12 at 0:41

Let's look at the question:

Find the domain of $f(x)=\dfrac{3x+1}{\sqrt{x^2+x-2}}$

Now to answer this, we need to know what the domain of a function is. Without getting into technical details, we can think of the domain as the values of $x$ that give us values of $y$. When answering a question to find the domain of a function, it is often more useful to try to find values of $x$ that do not give any value of $y$.

There are three basic kinds of functions to pay attention to: rational (fractions), sqaure roots, and logarithms. Here you have the first two. (I suspect that you haven't learned about logarithms, so we will ignore those for today.) For rational functions, the denominator cannot be zero. For square root functions, the radicand (stuff under the sqaure root symbol) cannot be negative. We can use these two facts to answer this problem.

First we set the denominator to zero:


You can square both sides of this equation and use the quadratic formula (or factor or complete the square) to find that $x = 1$ or $x = -2$ are solutions to the equation. This means that these two values of $x$ are not in the domain.

The next step is to determine which values of $x$ make the denominator of the original function negative. You can use the solutions to the above step to determine this. To do this, we first note that $x^2+x-2$ is zero when $x = 1$ or $x = -2$. So we know that every other value of $x$ will make the quadratic expression either positive or negative. So let's look at values of $x$ less than $-2$, between $-2$ and $1$ and greater than $1$. Which of these "intervals" make our quadratic function positive?

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So I have to find the values that make the radicand negative? – Austin Broussard Aug 7 '12 at 0:41
@AustinBroussard Exactly. Do you know how to do that? (Of course, from this point, it might be easier to find the values that make the radicand positive, but if you find one, you have the other almost automatically.) – Code-Guru Aug 7 '12 at 0:42
Not at all.. I'm going to guess this is pre-calculus stuff. Which I have never taken. I just went from Trig to Calculus. – Austin Broussard Aug 7 '12 at 0:43
Did you take algebra? Basically so called "pre-calculus" classes are just algebra. – Code-Guru Aug 7 '12 at 0:45
When $x=-3$ it's $4$ and then it goes up accordingly. $-1$ and $0$ yield $-2$. And $2$ gives $4$. Thus, the only negative values are between $-2$ and $1$. – Austin Broussard Aug 7 '12 at 0:53

Sometimes (like this one!) things get pretty clearer when we make algebra and geometry interplay. This kind of questions are attacked around here in senior year of junior high school (advanced mathematics) without any calculus at all, of course.

First, you must know that $\,x^2+x-2=(x+2)(x-1)\,$ is a parabola and, in fact, a "smiling" one, meaning: it opens upwards, and its roots or zeros are $\,x_{1,2}=-2,1\,$ (and for this we need either the quadratic formula or factoring the trinomial).

So we have a smiling parabola, and from the geometric picture we get at once this parabola in non-positive iff $\,-2\leq x\leq 1\,$ , so that it is positive whenever $\,x<-2\,\,\,or\,\,\,x>1\,$.

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I do not think that your rationalization step was necessary. Anyway, this is my solution. $$ x^2+x-2 >0 $$
or factorising LHS, you get, $$ (x+2)(x-1) >0 $$
or, $x$ lies in one of $(-\infty, -2)\text{ or }(1, \infty)$.

As for your solution, it was created unnecessarily long by you ( by rationalizing) in the first step and as for your question, in your last step alpha does not decide the domain or you can say this is because no restriction ( like if it is in denominator it cannot be equal to $0$) on the numerator term .

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