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How can I find equation for this function?

interesting function

the asymptotes on the sides can be for example at $x = -8$ and $x = 8.$

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    Yes, I meant expression. Function has only three vertical asymptotes. Everywhere else function is continuous. Let's say, that function intersects axe x in 2,3,4,5,6,7 and -2,-3,-4,-5,-6,-7. And also in +/- infinity it has limit to 02012-12-14

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Let's give it a try: since the function we look for $\,f(x)\,$ is even (according to the given graph), we can begin with $\,g(x)=(x^2-4)(x^2-9)(x^2-25)(x^2-36)(x^2-49)\,$ , and it has vertical asymptotes at $\,x=0\,,\,\pm 8\,$ , we can put

$k(x)=\frac{g(x)}{x^2(64-x^2)}$

(not $\,x^2-64\,$ in the denominator as we need, for example $\,f(x)\xrightarrow [x\to -8^-]{} -\infty\,$

and since we need

$\lim_{x\to \pm\infty}f(x)=0$

we need the denominator's degree higher than the numerator's but in a way as to be sensible to sign changes around $\,\pm\,8\,$, so for example

$f(x):=\frac{g(x)}{x^2(64-x^2)^{9}}$

Check carefully whether the above makes the cut.

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    For a straight like $\,x=x_0\,$ to be a vertical asymptote it is sufficient and necessary that *at least* either one of the one-sided limits of the function when $\,x\to x_0^+\,\,\,or\,\,\,x\to x_0^-\,$ is $\,\infty\,\,\,or\,\,\,-\infty\,$, which is precisely what happens in two of the three vertical asymptotes.2012-12-15