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# Find a formula for a function $f$ that satisfies the following conditions:$$\lim_{x \to \pm \infty} f(x) = 0, \lim_{x \to 0} f(x) = -\infty , f(2) = 0, \lim_{x \to 3^-} f(x) = \infty, \lim_{x \to 3^+} f(x) = -\infty$$

## $\frac{2-x}{x^{2}(x-3)}$

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##### Kristen K.

University of Michigan - Ann Arbor

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Okay, let's find a formula for a function F that satisfies the following conditions. The let's let's just draw a graph here and see if we can recognize this kind of thing. So the limit as X approaches positive or negative infinity. It's Sarah So we have two options here. Infinity means it's gonna go off this way or this way. Same thing on the other side. Mhm. You okay? The limit as X approaches zero is negative infinity. So this means on the left and the right As it approaches zero has to reach negative infinity which probably means that we have an X in our denominator and some kind of negative that gives us some clues as to what this function would look like. The limit as x equals 3 -3 is going to be infinity. So that's At 1 2, 3 here. So three from the left is going to be positive infinity. We And three from the right is going to be negative infinity. Which also tells me that we have an X -3 in the denominator. Uh huh. Okay, let's try to figure out what exactly this function could look like. So let's look at each of the clues We know that it has to approach zero Because it approaches zero. The degree of the denominator has got to be greater than the degree of the numerator. I know that because if you think of something like X over X cubed that simplifies to one over X squared or something. The nominator is so much bigger is going to grow so much faster, which means that no matter what the numerator is, It will outweigh that. So it's like one over 100,000, which is very close to zero, which means it will approach zero. So we know that the degree of the denominator has got to be greater than the degree of the numerator. We also know that there's got to be an X in the denominator And an X -3 in the denominator. That accounts for those acid totes. So let's just put a one on the top for now. If we put an X down here, let's think about this for a minute as we approach zero from the left, it's going to be one over negative zero point 00001 and one over a very small negative would give us negative infinity. So that works. But what about the positive side? Your 10.11 over a very small positive would give us positive infinity. So that doesn't work. There has to be some kind of exponent there. That makes both sides negative. So let's have a negative in front here, negative X squared. So let's think about this. If we have 0.00001 squared, that's an even smaller negative, even smaller number. If we have a negative there, it's one over a very small negative number which gives us the negative infinity we're looking for. So that works Now the X -3, Wolf. Okay, this should be fun. I'm going to add this negative end here like this. Okay, the X -3. Let's see what this equation might look like. So if we have our X -3, if we approach it from the right, a little bit more than three, then this will be the same for both terms. So we can kind of ignore that for now. If we have a little bit more than three, it's going to be won over a very small positive which would give us a positive infinity. If we have one over a very small negative, that would give us negative infinity. But it looks like this is flipped. So what can we do? Let's factor the negative out of here and see what happens. So now if I've three from the right 3.0000010 shoot, this is sorry about that. I should flip this that I should flip it so that it's 3 -1. And let me explain where I got that from. Remember that the original of X -3 would give us the opposite of what we're looking for with and ask them to. That looks like that on the right like that on the left. But this one if we plug in something slightly more than three, like from the right 3.1, it would be one over a very small negative, which would give us negative infinity and the opposite on the right hand side. And remember I'm kind of ignoring this negative X squared Because it's there on either side of the limit so I can kind of cancel it. It doesn't matter as much. This is the one I'm really looking at, the 3 -6. So let's see if that satisfies all the conditions. I think I think it might. So the degree of the denominator is greater than the degree of the numerator. That's true. We plug in a huge number down here. It's going to be won over that. So like one over 1000 is going to be very close to zero. So the function should approach zero the way we want it to, which satisfies the condition of the limit as X approaches as X approaches plus or minus infinity is zero. And then the second part about it approaching um negative as it approached X approaches zero equals negative infinity. That also works here. If we plug in a little negative or little positive, we're going to find that it's negative infinity because it's one over a negative and this negative out front helps that happen. If it was if there wasn't a negative out here, it would point upwards instead on both sides. Now, finally, the three -X. So this satisfies the condition of As X approaches three from the left this side, It will approach infinity. Think about like 2.99999993 -2.99999 is a very small positive. One over a very small positive is positive infinity. On the other side, if we have 3.01, we have one over a very small positive, which gives us sorry, Yeah, one over a very small negative, which gives us negative infinity. So this functions fulfills all those conditions. And you can yeah, prove it to yourself, plug in some numbers and see how it goes.

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The University of Alabama

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