Determine if:
f(x) = 1/ x-2 and g(x)= 2x +1/x are inverse functions by computing their composistions.
Hi,
you get the inverse function if you change the variables.
If $\displaystyle f:y=\frac{1}{x-2}$ then the inverse funtion of f
is $\displaystyle f^{-1}:x=\frac{1}{y-2}$
$\displaystyle (y-2)\cdot x=1$
$\displaystyle y-2=\frac{1}{x}$
$\displaystyle y=\frac{1}{x}+2$
$\displaystyle y=\frac{1}{x}+\frac{2x}{x}$
$\displaystyle y=\frac{1+2x}{x}$ and that's your function g. So $\displaystyle g = f^{-1}$
Now proof that the inverse function of g is actually the function f. I leave this to you.
EB
Please use parenthesis! f(x) = 1/(x-2) not f(x) = (1/x) - 2 and g(x) = (2x + 1)/x, not g(x) = 2x + (1/x).
$\displaystyle f(g(x)) = f \left ( \frac{2x+1}{x} \right ) = \frac{1}{\left ( \frac{2x+1}{x} \right ) - 2}$ Multiply top and bottom by x.
= $\displaystyle \frac{1}{\left ( \frac{2x+1}{x} \right ) - 2} \cdot \frac{x}{x}$ = $\displaystyle \frac{1 \cdot x}{\left ( \frac{2x+1}{x} \right )x - 2x}$ = $\displaystyle \frac{x}{2x+1 - 2x}$
= $\displaystyle \frac{x}{1} = x$ (Check!)
Again:
$\displaystyle g(f(x)) = g \left ( \frac{1}{x-2} \right ) = \frac{2 \left ( \frac{1}{x-2} \right ) + 1}{\left ( \frac{1}{x-2} \right )}$ Multiply top and bottom by x - 2.
= $\displaystyle \frac{2 \left ( \frac{1}{x-2} \right ) + 1}{\left ( \frac{1}{x-2} \right )} \cdot \frac{x-2}{x-2}$ = $\displaystyle \frac{2 \left ( \frac{1}{x-2} \right ) (x-2) + (x-2)}{\left ( \frac{1}{x-2} \right ) (x-2)}$
= $\displaystyle \frac{2 + (x - 2)}{1} = 2 + x - 2 = x$ (Check!)
So f(g(x)) = x and g(f(x)) = x. Thus f and g are inverses.
-Dan