How can I sketch the reciprocal of a function with poles at x=-2 and x=2?

AI Thread Summary
To sketch the reciprocal of a function with poles at x=-2 and x=2, recognize that the reciprocal will have roots at these points. The original function is symmetric about the y-axis, which means the reciprocal will also exhibit this symmetry. At x=0, the function's value is slightly above -1, indicating the reciprocal will be slightly below -1. As x approaches 2, the reciprocal graph will rise to y=0 and continue increasing toward positive infinity. This behavior suggests that the reciprocal graph resembles a quadratic function, particularly similar to the graph of -sec x, leading to a reciprocal resembling -cos x.
danago
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Given the following graph:
http://img245.imageshack.us/img245/2395/scan0001ou4.gif

How can i sketch the reciprocal of that function? There are poles at x=-2 and x=2, so it means its reciprocal will have roots at -2 and 2 right? But that's not really enough information to compose a full sketch.

How should i go about doing this?

Thanks,
Dan.
 
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It's enough for a rough sketch. You can see that the function is symmetric about the y-axis so its reciprocal will be. You can see that at x= 0, the function has value just a little larger than -1 so its reciprocal will have value just a little less than -1. The reciprocal graph will start at x= 0, y= a little less than -1, rise to x= 2, y= 0, then continue increasing as x goes to + infinity. Use symmetry to get the graph for negative x.
 
replace each y with 1/y for all x.
eg. where y is tending to infinity, it should tend to 0.
 
So it will be similar to a quadratic graph?
 
danago said:
So it will be similar to a quadratic graph?
the middle portion is very similar to graph of -sec x, so its reciproca would be similar to -cos x
 
1/big = small

1/small = big
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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