inverse funtions

**robinpatrick** · October 23rd 2006, 08:56 PM

Determine if:
f(x) = 1/ x-2 and g(x)= 2x +1/x are inverse functions by computing their composistions.

**earboth** · October 24th 2006, 01:22 AM

Originally Posted by robinpatrick

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 $f:y=\frac{1}{x-2}$ then the inverse funtion of f
is $f^{-1}:x=\frac{1}{y-2}$
$(y-2)\cdot x=1$
$y-2=\frac{1}{x}$
$y=\frac{1}{x}+2$
$y=\frac{1}{x}+\frac{2x}{x}$
$y=\frac{1+2x}{x}$ and that's your function g. So $g = f^{-1}$

Now proof that the inverse function of g is actually the function f. I leave this to you.

EB

**topsquark** · October 24th 2006, 06:00 AM

Originally Posted by robinpatrick

Determine if:
f(x) = 1/ x-2 and g(x)= 2x +1/x are inverse functions by computing their composistions.

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).

$f(g(x)) = f \left ( \frac{2x+1}{x} \right ) = \frac{1}{\left ( \frac{2x+1}{x} \right ) - 2}$ Multiply top and bottom by x.

= $\frac{1}{\left ( \frac{2x+1}{x} \right ) - 2} \cdot \frac{x}{x}$ = $\frac{1 \cdot x}{\left ( \frac{2x+1}{x} \right )x - 2x}$ = $\frac{x}{2x+1 - 2x}$

= $\frac{x}{1} = x$ (Check!)

Again:
$g(f(x)) = g \left ( \frac{1}{x-2} \right ) = \frac{2 \left ( \frac{1}{x-2} \right ) + 1}{\left ( \frac{1}{x-2} \right )}$ Multiply top and bottom by x - 2.

= $\frac{2 \left ( \frac{1}{x-2} \right ) + 1}{\left ( \frac{1}{x-2} \right )} \cdot \frac{x-2}{x-2}$ = $\frac{2 \left ( \frac{1}{x-2} \right ) (x-2) + (x-2)}{\left ( \frac{1}{x-2} \right ) (x-2)}$

= $\frac{2 + (x - 2)}{1} = 2 + x - 2 = x$ (Check!)

So f(g(x)) = x and g(f(x)) = x. Thus f and g are inverses.

-Dan

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