MHB Solve x^2/(x+3) < 9/(x+3): -2 < x < 1/2, x > 3

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The discussion revolves around solving the inequality x^2/(x+3) < 9/(x+3). The initial approach involved cross-multiplying and simplifying to get x^2 - 9, which factors to (x+3)(x-3), leading to an interval analysis. However, the book's solution indicates the intervals -2 < x < 1/2 and x > 3, which the participants find confusing. They clarify that x = -3 is a hole in the graph rather than a vertical asymptote, emphasizing the importance of not removing critical points in the solution process. Ultimately, the participants express skepticism about the book's answer while discussing the implications for graphing the function.
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Find the solution set for x^2/(x+3) < 9/(x+3)

So I moved the term 9/(x+3) over to the left side and cross-multiplied the two fractions. Then, I simplified to get x^2-9 (because the x+3 cancel out across the fraction bar). I got x^2-9, which factors to (x+3)(x-3). Then, I created an interval table using these key values, ending with the answer -3 < x < 3.

However, the answer in the book states that it is -2 < x < 1/2 and x > 3. What did I do wrong? How did they get the -2 and 1/2 values?

Thanks!
 
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This is how I would solve the problem:

$$\frac{x^2}{x+3}<\frac{9}{x+3}$$

$$\frac{x^2}{x+3}-\frac{9}{x+3}<0$$

$$\frac{(x+3)(x-3)}{x+3}<0$$

We don't want to remove the singularity here, as this will remove a critical number. We have two such critical numbers:

$$x\in\{-3,3\}$$

Both if which need to be excluded from the solution because of division by zero in the case of $x=-3$ and because the inequality is strict.

So, we have 3 intervals to consider:

$$(-\infty,-3),\,(-3,3),\,(3,\infty)$$

Pick a test value from each interval, and test the sign of the expression on the left, and keep those for which the expression is negative...what do you find?
 
MarkFL said:
This is how I would solve the problem:

$$\frac{x^2}{x+3}<\frac{9}{x+3}$$

$$\frac{x^2}{x+3}-\frac{9}{x+3}<0$$

$$\frac{(x+3)(x-3)}{x+3}<0$$

We don't want to remove the singularity here, as this will remove a critical number. We have two such critical numbers:

$$x\in\{-3,3\}$$

Both if which need to be excluded from the solution because of division by zero in the case of $x=-3$ and because the inequality is strict.

So, we have 3 intervals to consider:

$$(-\infty,-3),\,(-3,3),\,(3,\infty)$$

Pick a test value from each interval, and test the sign of the expression on the left, and keep those for which the expression is negative...what do you find?

So I tried that and got that it's negative for the intervals x<-3 and -3 < x < 3. But that's still different from the answer in the book? I'm not sure where they'd even get the -2 and 1/2 from.
 
eleventhxhour said:
So I tried that and got that it's negative for the intervals x<-3 and -3 < x < 3. But that's still different from the answer in the book? I'm not sure where they'd even get the -2 and 1/2 from.
(shrugs) The book is wrong.

-Dan
 
topsquark said:
(shrugs) The book is wrong.

-Dan

Alright, thanks!

So the book also asked to graph it. I know there'd be vertical asymptote at x = -3, but how would you figure out where the horizontal asymptote is (or if there even is one)?
 
eleventhxhour said:
...So the book also asked to graph it. I know there'd be vertical asymptote at x = -3, but how would you figure out where the horizontal asymptote is (or if there even is one)?

You won't have a vertical asymptote at $x=-3$, you will have a hole in the graph instead, because the singularity is removable.

It will be identical to the graph of $y=x-3$, except with the hole at $(-3,-6)$.
 
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