To satisfy the inequality, sketch the following function by hand and shade the graph regions.

Hi chibi, where's the inequality?

Also, I want to make sure the equation is written correctly:

\(f(x)=\frac{1}{x^2}-9\) or \(f(x)=\frac{1}{x^2-9}\)

If it's the latter, notice that the denominator is a *difference of squares*

\(f(x)=\frac{1}{\left(x+3\right)\left(x-3\right)}\)

Not sure if that helps, but you need to verify if the equation is correct, and if you wrote the <, > symbol wrong.

Hi @bio_man, my apologies, the inequality is 1/x^2-9 > 0. Yes, the second equation is correct.

I believe the answer looks something like this:



Let me know what you think

Hi @bio_man, thank you, this is correct. Although the question is just asking for a sketch, I would like a more accurate graph. What would be the list of points for this in particular?

Good question. So once you find the asymptotes, you know the behavior of the graph will be unusual close to these numbers.

What I like to do is find out what happens between -3 and 3, and in-between.

Choose a number close to negative 3 (but less than it), like -2.9999. Plug it into the equation, see what you get. This is sort of like taking the limit of the function, if you've ever done that before

\(f(-2.9999)=\frac{1}{\left(x+3\right)\left(x-3\right)}=-1666.69\)

Notice how the output is so negative, which suggests that it'll go to negative infinity the closer we get to -3.

Do the same for a number close to 3, but not quite 3:

\(f(2.9999)=\frac{1}{\left(x+3\right)\left(x-3\right)}=-1666.69\)

Again, you get the same output, something that's very negative. Also, in-between -3 and 3 is 0, sub. that in too:

\(f(0)=\frac{1}{\left(x+3\right)\left(x-3\right)}=\frac{-1}{9}\)

Now you get a sense of how this equation behaves between its asymptotes. You carry on doing this to the left of -3 and the right of 3, then you'll get a sense of how it behaves overall.

If you want a more accurate graph, of course you could do a table of values.

Hi @bio_man, thanks for that. Finished with this question