f(x)= x/(x^2-4)
The domain for this function is restricted to -2<x<2
Can anyone help with this? Thanks!
$\displaystyle x = \frac{y}{y^2-4}$
$\displaystyle xy^2 - 4x = y$
$\displaystyle xy^2 - y - 4x = 0$
$\displaystyle y = \frac{1 \pm \sqrt{1 - 4(x)(-4x)}}{2x}$
$\displaystyle y = \frac{1 \pm \sqrt{1 + 16x^2}}{2x}$
I'll leave you to determine whether the numerator should be
$\displaystyle 1 + \sqrt{1 + 16x^2}$ or $\displaystyle 1 - \sqrt{1 + 16x^2}$
Well I got that, but I'm unsure as to which one it is. I have a feeling that its the second one (the one with the negative sign) because when I made data tables, the x and y values were switched, but there has to be an easier method. Can anyone help?
And thanks skeeter!
A simple approach:
Get a simple convenient point lying on $\displaystyle y = f(x)$. Eg. $\displaystyle x = 1 \Rightarrow y = -\frac{1}{3}$. So $\displaystyle \left( 1, ~ - \frac{1}{3}\right)$ is a simple point lying on $\displaystyle y = f(x)$.
This means that the point $\displaystyle \left(- \frac{1}{3}, ~ 1\right)$ MUST lie on $\displaystyle y = f^{-1}(x)$. Which choice for the numerator makes this happen?
Hello, MegaVortex7!
I had to baby-talk my way through this one . . .
$\displaystyle f(x)\:=\:\frac{x}{x^2-4}$
The domain for this function is restricted to $\displaystyle -2\:<\:x\:<\:2$
Find the inverse function, $\displaystyle f^{-1}(x).$
To visualize the effect of the domain, I graphed the function.Code:: | : :* | : : | : : * | : : * | : : * | : : * | : - - + - - - - - * - - - - - + - - -2: | * :2 : | * : : | * : : | * : : | : : | *: : | :
Then the inverse function looks like this:To find the inverse . . .Code:| - - - - - - - - + - - - - - - - - * |2 * | * | * | *| | - - - - - - - - - * - - - - - - - - - | |* | * | * | * | * - - - - - - - - + - - - - - - - - - - |-2
Replace $\displaystyle f(x)$ with $\displaystyle y\!:\;\;y \;=\;\frac{x}{x^2-1}$
Switch $\displaystyle x$'s and $\displaystyle y$'s: .$\displaystyle x \;=\;\frac{y}{y^2-1}$
Solve for $\displaystyle y\!:$
. . $\displaystyle xy^2 - 4x \:=\:y \quad\Rightarrow\quad xy^2 - y -4x \:=\:0$
Quadratic Formula: .$\displaystyle y \;=\;\frac{1 \pm \sqrt{1 + 16x^2}}{2x}$
If the inverse is to resemble its graph above,
. . (as x → +∞, y → -2)
. . we must have: .$\displaystyle \boxed{f^{-1}(x) \;=\;\frac{1 - \sqrt{1 + 16x^2}}{2x}}$
Nice and simple but then the OP has to be sophisticated enough to realise that you make the choice that gives an indeterminant form. This choice should be obvious since the other choice gives something undefined. Nevertheless, I'd predict hand wringing and angst with this choice .......