Finding a function with given conditions

Question

I want to find the equation of a function $f(x)$ such that the following conditions are met:

$(i)\ f(3) = 0, f(0) =1$.

$(ii)\ f(x)$ is an even function.

$(iii)\ f(x)$ has vertical asymptodes at $x=\pm 4$.

$(iv)\ f(x)$ has horizontal asymptode at $ y=2$.

Now since the function is even, I get $x=-3$ as another root and then I can graph the function. But how do I find its equation? Any hints? Thank you.

In general is there some method or software to find the equation of a function given its graph?

Answer 1

We use building blocks having one or more of the wanted properties and put them together conveniently so that the final function $f$ has all wanted properties.

• We start with (iii) and take as function with asymptote at $x=\pm 4$ the rational function \begin{align*} x\to\frac{1}{(x-4)(x+4)}=\frac{1}{x^2-16}\tag{1} \end{align*}

  which is already even.

To get a zero at $x=3$ we multiply with $x-3$ and to keep the function even we multiply with $(x-3)(x+3)=x^2-9$.

• We obtain \begin{align*} x\to\frac{x^2-9}{x^2-16}\tag{2} \end{align*}

Note that $\frac{x^2-9}{x^2-16}$ evaluated at $x=0$ is $\frac{9}{16}$.

• We multiply the function (2) with $\frac{16}{9}$ and obtain with \begin{align*} x\to\frac{16}{9}\cdot\frac{x^2-9}{x^2-16}\tag{3} \end{align*} an even function which fulfills (i) and (ii).

Finally we extend (3) to get a function with horizontal asymptote at $y=2$. Note that multiplying the function with $e^{-x^2}$ gives \begin{align*} \lim_{x\to\pm\infty}\frac{x^2-9}{x^2-16}e^{-x^2}=0 \end{align*}

Since we also have \begin{align*} \lim_{x\to\pm\infty}\frac{x^2-9}{x^2-16}=1 \end{align*}

we use the Ansatz \begin{align*} f(x)=\frac{16}{9}\cdot\frac{x^2-9}{x^2-16}\left(ae^{-x^2}+b\right) \end{align*} solve \begin{align*} f(0)&=a+b=1\\ \lim_{x\to\pm\infty}f(x)&=\frac{16}{9}b=2 \end{align*} and find $a=-\frac{1}{8}$ and $b=\frac{9}{8}$.

• We finally obtain a function
  \begin{align*} &\color{blue}{f:\mathbb{R}\to\mathbb{R}\setminus\{\pm 4\}}\\ &\color{blue}{f(x)=\frac{2}{9}\cdot\frac{x^2-9}{x^2-16}\left(-e^{-x^2}+9\right)} \end{align*} which fulfills (i) to (iv).

Note: To find an even function is usually not a problem. When we have a function $g$ fulfilling all other properties we can take $f(x)=\frac{1}{2}\left(g(x)+g(-x)\right)$ to get an even function with the wanted properties.

Answer 2

I am assuming you want to find a rather than all functions that satisfy that property. You can approach this with a sort of an ad hoc method.

First lets find some function $f(x)$ that satisfies $(i)$. This should be easy enough, the polynomial $x(x-3)$ does the job. (This i incorrect as I misread the condition to be $f(0) = 0$, this is fixed in the edit below)

Now for $(ii)$, we want $f(x) = f(-x)$. Notice squaring a number makes its sign redundant so we try incorporating that into our function to get $x^2(x^2-9)$. (always check back to see you havent lost $(i)$)

$(iii)$ requires us to divide by $0$ when $x = \pm 4$. So let $$f(x) = \frac{x^2(x^2-9)}{(x-4)(x+4)} = \frac{x^2(x^2-9)}{(x^2-16)}$$

Notice how this still has properties $(i)(ii)$

Finally, we need the horizontal asymptotic condition. Notice right now that $\lim_{x \rightarrow \infty} f(x) = \infty$ since the numerator is of degree $4$ while the denominator is of degree $2$. We can first let $f(x)$ converge when $x$ goes to infinity simply by squaring the denominator. This results in $$f(x) = \frac{x^2(x^2-9)}{(x^2-16)^2}$$

This gives us a horizontal asymptote of $y=1$. Multiplying our function by $2$ then fixes this $$f(x) = 2\frac{x^2(x^2-9)}{(x^2-16)^2}$$ we are done.

EDIT (Sorry for the error) we remove the $x^2$ and the square on the denominator to get $\frac{2(x^2-9)}{x^2 - 16}$ Notice at $x = 0$ this evaluates to $9/8$. We want to add a term that disappears at $x = \pm 3$ and as $x \rightarrow \infty $ and evaluates to $-1/8$ when $x = 0$. Following similar line of reasoning as above we get $\frac{(x^2 - 9)16^2}{9(8)(x^2 - 16)^2}$ so our function becomes $$f(x) = \frac{2(x^2-9)}{x^2 - 16} + \frac{(x^2 - 9)16^2}{9(8)(x^2 - 16)^2} = (\frac{x^2-9}{x^2 -16})(2+ \frac{32}{9(x^2 -16)})$$