1. ## Checking for inverses

OK so i attached a copy of my question. I have MANY questions for this assignment set so I am not just trying to get somebody to do my work for me. I dont understand how to compose this function (either in g of inverse g of x or inverse g of g of x) because each function has an x in the denominator.

2. To find the inverse of

$\displaystyle g = 1-\frac{1}{x-1}$

swap g and x then solve for g.

$\displaystyle x = 1-\frac{1}{g-1}$

$\displaystyle x -1 = -\frac{1}{g-1}$

$\displaystyle 1 -x = \frac{1}{g-1}$

Flip both sides over

$\displaystyle \frac{1}{1 -x} = \frac{g-1}{1}$

$\displaystyle \frac{1}{1 -x} = g-1$

$\displaystyle 1+\frac{1}{1 -x} = g$

$\displaystyle g = 1+\frac{1}{1 -x}$

3. Originally Posted by komplex85
I dont understand how to compose this function...because each function has an x in the denominator.
When composing functions, the particular location of the "x" is irrelevant. When evaluating f(x) at g(x) (that is, when "finding f(g(x))"), plug the expression for "g(x)" in for the "x" in "f(x)".

. . . . .$\displaystyle g(x)\, =\, 1\, -\, \frac{1}{x\, -\, 1}$

. . . . .$\displaystyle g^{-1}(x)\, =\, 1\, +\, \frac{1}{1\, -\, x}$

. . . . .$\displaystyle g(g^{-1}(x))\, =\, 1\, -\, \frac{1}{g^{-1}(x)\, -\, 1}$

. . . . . . . .$\displaystyle =\, 1\, -\, \frac{1}{\left(1\, +\, \frac{1}{1\, -\, x}\right)\, -\, 1}$

. . . . . . . .$\displaystyle =\, 1\, -\, \frac{1}{1\, -\, 1\, +\, \left(\frac{1}{1\, -\, x}\right)}$

. . . . . . . .$\displaystyle =\, 1\, -\, \frac{1}{\left(\frac{1}{1\, -\, x}\right)}$

. . . . . . . .$\displaystyle =\, 1\, -\, (1\, -\, x)$

. . . . . . . .$\displaystyle =\, 1\, -\, 1\, +\, x$

...and so forth.