chengeto said:
What is the inverse of the function : y=3x-(1/2x) on restricted domain D={x|x<0}. Verify that f^-1(f(x)) = x for all x in the restricted domain.
I will guess that you mean the function to be y = 3x - 1/(2x).
You appear to have started by multiplying through by the denominator 2x, and then attempted to solve the equation for "x=".
. . . . .\(\displaystyle y\, =\, 3x\, -\, \frac{1}{2x}\)
. . . . .\(\displaystyle 2xy\, =\, 6x^2\, -\, 1\)
. . . . .\(\displaystyle 0\, =\, 6x^2\, -\, 2xy\, -\, 1\)
This is a quadratic in "x", and you need to solve for x, so:
. . . . .\(\displaystyle x\, =\, \frac{-(-2y)\, \pm\, \sqrt{(-2y)^2\, -\, 4(6)(-1)}}{2(6)}\)
. . . . .\(\displaystyle x\, =\, \frac{2y\, \pm\, \sqrt{4y^2\, +\, 24}}{12}\)
. . . . .\(\displaystyle x\, =\, \frac{2y\, \pm\, 2 \sqrt{y^2\, +\, 6}}{12}\)
. . . . .\(\displaystyle x\, =\, \frac{y\, \pm\, \sqrt{y^2\, +\, 6}}{6}\)
And then you took the negative root (since x < 0) and reversed the variables to get your inverse. So this part looks good.
To confirm the inverse, you need to compose the two functions. So plug the original formula in for "x" in your inverse:
. . .\(\displaystyle f^{-1}(f(x))\, =\, \frac{\left(3x\, -\, \frac{1}{2x}\right)\, -\, \sqrt{\left(3x\, -\, \frac{1}{2x}\right)^2\, +\, 6}}{6}\)
. . . . .\(\displaystyle =\, \frac{\frac{6x^2\, -\, 1}{2x}\, -\, \sqrt{9x^2\, -\, 3\, +\, \frac{1}{4x^2}\, +\, 6}}{6}\)
. . . . .\(\displaystyle =\, \frac{\frac{6x^2\, -\, 1}{2x}\, -\, \sqrt{9x^2\, +\, 3\, +\, \frac{1}{4x^2}}}{6}\)
. . . . .\(\displaystyle =\, \frac{\frac{6x^2\, -\, 1}{2x}\, -\, \sqrt{\left(3x\, +\, \frac{1}{2x}\right)^2}}{6}\)
. . . . .\(\displaystyle =\, \frac{\frac{6x^2\, -\, 1}{2x}\, -\, \sqrt{\left(\frac{6x^2\, +\, 1}{2x}\right)^2}}{6}\)
Keeping in mind that x < 0, so 2x < 0, so the square root of the square of 2x is actually |2x| = -(2x):
. . . . .\(\displaystyle =\, \frac{\frac{6x^2\, -\, 1}{2x}\, -\, \frac{6x^2\, +\, 1}{-2x}}{6}\)
. . . . .\(\displaystyle =\, \frac{\frac{6x^2\, -\, 1}{2x}\, +\, \frac{6x^2\, +\, 1}{2x}}{6}\)
. . . . .\(\displaystyle =\, \left(\frac{6x^2\, -\, 1\, +\, 6x^2\, +\, 1}{2x}\right) \left(\frac{1}{6}\right)\)
. . . . .\(\displaystyle =\, \left(\frac{12x^2}{2x}\right) \left(\frac{1}{6}\right)\)
I'll bet you can take it from there! :wink:
