Given the curve $$y={1 \over{x^2-1}}-{1\over {x^2}}$$ Find reasonable approximations for the intersections of this curve with the straight line $y=Ax$ (a) when $A$ is a very small positive number (b) when $A$ is a very large positive number.
This is all the information I was given, but I am confused as to how small is small? Also, what does reasonable mean?
What would be your answer to this question? How should I interpret it?
Note: The question also asks one to sketch $y$.
On
These questions can be hard when you have to interpret it, but usually the best way to go is to see what happens if you would take $A$ a large number (say $1000$), and then double it. Then check to see what exactly changes, and where there some things that are negligible. In this case, $y$ goes to infinity as $x\rightarrow 1^+$, so that $y$ resembles a straight vertical line more and more as you approach $x = 1$ from the right. What can you say about the intersection of a line $Ax$ with a vertical line at $x = 1$?
A same strategy can also be done for $A$ is small.
An alternative approach to sketching your rational function is to do some algebra to convert the problem into the equation
$$x^5-x^3-\frac{1}{A}=0.$$
Note that none of $-1,0,1$ are solutions to this modified equation, so we don't need to enforce that requirement explicitly.
This function is not hard to sketch. In fact, you can find its relative maximum explicitly. The maximum value turns out to be $M=\frac{6 \sqrt{3/5}}{25}-\frac{1}{A}$. When $A$ is small, $M<0$, and there is only one root, which is on $(1,\infty)$. At a single value of $A$ in the middle, you have $M=0$, in which case there are two roots, one on $(1,\infty)$ and another at the maximum, which is at $-\sqrt{3/5}$. When $A$ is large, $M>0$, and there are three roots, two on $(-1,0)$ and opposite sides of the maximum, and again one on $(1,\infty)$.
For small $A$, the single root will need to be very large, so that $x^5 \gg x^3$. So you can drop the smaller term and solve.
For large $A$, the root on $(1,\infty)$ will be just slightly larger than $1$ (so that the solution to this equation is near the solution to $x^5-x^3=0$). There $x^5-x^3$ can be linearly approximated by $2(x-1)$, so you can approximate by the solution to $2(x-1)=\frac{1}{A}$, which is $x=1+\frac{1}{2A}$. The situation is essentially identical at $-1$.
At $0$ you can't linearize and get a reasonable answer. This is because the root there is a triple root. Instead, you can drop the $x^5$ term since $x^5 \ll x^3$ near $0$, and get $-x^3-\frac{1}{A}=0$ as your approximate equation.