Determine the point of intersection of the following two rational functions, algebraically.
r(x)=$\displaystyle x/(3x^2+3)$ s(x)=$\displaystyle 1/(2x^2-8)$
Thank you
The point of intersection is going to be defined by $\displaystyle r(x) = s(x)$ for some values of x. So:
$\displaystyle \frac{x}{3x^2 + 3} = \frac{1}{2x^2 - 8}$
Cross multiply (or, more precisely, multiply both sides by $\displaystyle (3x^2 + 3)(2x^2 - 8)$)
$\displaystyle x(2x^2 - 8) = 3x^2 + 3$
$\displaystyle 2x^3 - 3x^2 - 8x - 3 = 0$
You can use the rational root theorem to get the solutions. I get $\displaystyle x = -1, -1/2, 3$.
-Dan
First, note where the two functions might not be defined: s(x) is undefined for $\displaystyle x=\pm2$, so those points cannot be intersections.
Then we set r(x)=s(x):
$\displaystyle \frac{x}{3x^2+3}=\frac{1}{2x^2-8}$
Multiply both sides by the product of the denominators and cancel:
$\displaystyle x(2x^2-8)=1(3x^2+3)$
Expand, and use algebra to move all terms to one side:
$\displaystyle 2x^3-3x^2-8x-3=0$
This is a cubic equation, and there are methods of solving it. However, once you find one root, $\displaystyle r_1$, you can factor out the term $\displaystyle (x-r_1)$ from the polynomial to obtain a quadratic which will tell you if there are two other real roots, and what they are (if they exist). Using the Rational Root Theorem can help you find that first root.
--Kevin C.