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Good day! I encounter a problem on rational function wherein I almost satisfy all the given requirements to derive the rational function except the last one. Any idea would be of great help, Thanks!

Find the function satisfying the following:

  1. $f(3) = 0$
  2. $f(x) = f(-x)$
  3. Vertical Asymptotes $x = 4$ and $x = -4$
  4. Horizontal Asymptotes $y = 2$
  5. $f(0) = 1$
user463383
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rosa
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  • Well, what have you tried? – Simply Beautiful Art Jul 16 '17 at 11:36
  • @SimplyBeautifulArt Took my line? – Parcly Taxel Jul 16 '17 at 11:37
  • Is that supposed to say $x=-44$ or $x=-4$ for the vertical asymptote? – Simply Beautiful Art Jul 16 '17 at 11:37
  • @SimplyBeautifulArt x = -4 as one of the vertical asymptotes. I already tried to form the equation, satisfying the first four requirements, but the fifth one give me the trouble for it does not satisfy my formed equation. I think its the trickiest part... – rosa Jul 16 '17 at 11:39
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    It would help a great deal if you edited your question to include your efforts so that we could assist you better. – Simply Beautiful Art Jul 16 '17 at 11:43
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    From (1) $(x-3)$ is a factor. From (2) $(x+3)$ is also a factor. From (3) $(x-4)(x+4)=x^2-16$ is a factor of the denominator. So far $\frac{x^2-9}{x^2-16}$ is a factor. Now, this factor tends to $1$ as $x\to\pm\infty$. Also, evaluating it at $0$ gives $\frac{9}{16}$. That means that the other factors must tend to $2$ as $x\to\pm\infty$ and at zero it must be equal to $\frac{16}{9}$. So, the other factor could be $\frac{16}{9}+(2-\frac{16}{9})\frac{x^2}{x^2-16}$. The problem says "the function". This suggests a unique solution. But there are many solutions. – user463383 Jul 16 '17 at 11:44
  • In conclusion, $\frac{x^2-9}{x^2-16}\left(\frac{16}{9}+(2-\frac{16}{9})\frac{x^2}{x^2-16}\right)$ is a rational function satisfying all requirements. – user463383 Jul 16 '17 at 11:48
  • @user463383 tnx a lot,, I understand ur idea :) – rosa Jul 16 '17 at 13:14

1 Answers1

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Hint. It is easy to verify that the following rational map has properties 2. 3. 4. $$h(x)=\frac{2x^2}{(x^2-16)}.$$ Now let us find a rational function $g$ such that $f=h+g$ has all the requested properties. For $g$ we need an even rational functions which has no vertical asymptotes and goes to $0$ as $x\to \pm \infty$. For example we can try something like $$g(x)=\frac{A}{1+x^2}+\frac{B}{2+x^2}$$ where the constants $A$ and $B$ have to be determined such that $f(3)=0$ and $f(0)=1$. So it remains to solve the linear system $$\begin{cases} \frac{A}{10}+\frac{B}{11}=0-h(3)\\ A+\frac{B}{2}=1-h(0) \end{cases}$$ which has a unique solution (the same procedure works for the more general situation where $f(3)=a$ and $f(0)=b$).

P.S. We expect "several" solutions for this problem.

Robert Z
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    Any further doubt? – Robert Z Jul 16 '17 at 12:15
  • I solved ur system of equation sir,, and eventually I come up with $A = \frac{-7325}{252}$ and $B = \frac{30371}{504}$ but as I evaluate f(3) = 0 and f(0) = 17/16 which is somewhat closed to 1. Thanks to your idea sir :) – rosa Jul 16 '17 at 13:18
  • @rosa Finally the solutions is $A = -1840/63$, $B = 3806/63$ and the $f(3)=0$ and $f(0)=1$. – Robert Z Jul 16 '17 at 14:52