Question 825489
I'm guessing that is not what you posted:
{{{f(x) = 1/2-x}}} 
but
{{{f(x) = 1/(2-x)}}}
because it makes the problem more interesting. If I am wrong, then I hope you will be able to figure out the answer to the real problem from the solution I show you below. If I am right, then please put parentheses around numerators and denominators which are not just a positive number or variable. For example:
f(x) = 1/(2-x)<br>
(fog)(x) means f(g(x)). And f(g(x)) represents the output of f when g(x) is used as input. So:
(fog)(x) = {{{f(g(x)) = f(2/x) = 1/(2-(2/x))}}}<br>
The domain of (fog)(x) is the set of all the possible values of x which can be used. Often the domain is all real numbers. But in this case, with an x in the denominator of {{{2/x}}} and in the denominator of {{{1/(2-(2/x))}}}, we have to make sure that neither denominator becomes a zero, <i>which we can <u>never</u> allow to happen!</i>. So let's see what values of x would make each denominator equal to zero:
For {{{2/x}}}, the denominator would be zero only if x was zero, too. For {{{1/(2-(2/x))}}}, the denominator would be zero only if x was 1. (If you cannot see this, then set the denominator equal to zero:
{{{2-(2/x) = 0}}}
and solve for x. You should get x = 1.<br>
Remember, we have found what numbers x <u>cannot</u> be. The domain is the set of numbers x can be. So the domain for (fog)(x) is: All real numbers <i>except</i> 0 and 1.