document.write( "Question 825489: find the domain of (fog)x given that f(x) =1/2-x and g(x) =2/x. \n" ); document.write( "

Algebra.Com's Answer #497384 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
I'm guessing that is not what you posted:
\n" ); document.write( " $\"f%28x%29+=+1%2F2-x\"$
\n" ); document.write( "but
\n" ); document.write( " $\"f%28x%29+=+1%2F%282-x%29\"$
\n" ); document.write( "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:
\n" ); document.write( "f(x) = 1/(2-x)

\n" ); document.write( "(fog)(x) means f(g(x)). And f(g(x)) represents the output of f when g(x) is used as input. So:
\n" ); document.write( "(fog)(x) = $\"f%28g%28x%29%29+=+f%282%2Fx%29+=+1%2F%282-%282%2Fx%29%29\"$

\n" ); document.write( "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 $\"2%2Fx\"$ and in the denominator of $\"1%2F%282-%282%2Fx%29%29\"$ , we have to make sure that neither denominator becomes a zero, which we can never allow to happen!. So let's see what values of x would make each denominator equal to zero:
\n" ); document.write( "For $\"2%2Fx\"$ , the denominator would be zero only if x was zero, too. For $\"1%2F%282-%282%2Fx%29%29\"$ , the denominator would be zero only if x was 1. (If you cannot see this, then set the denominator equal to zero:
\n" ); document.write( " $\"2-%282%2Fx%29+=+0\"$
\n" ); document.write( "and solve for x. You should get x = 1.

\n" ); document.write( "Remember, we have found what numbers x cannot be. The domain is the set of numbers x can be. So the domain for (fog)(x) is: All real numbers except 0 and 1. \n" ); document.write( "

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