document.write( "Question 155654: I am having touble with this F of G and G of F.
\n" ); document.write( "A) Determine the domain and range of a) F(x)= 2/(x^2+1) and b) g(x)=Square root of x-3\r
\n" ); document.write( "\n" ); document.write( "B) Find the formula for f(g(x)) and the domain of f(g(x))\r
\n" ); document.write( "\n" ); document.write( "C) Find the formula for g(f(x)) and the domain of g(f(x)).\r
\n" ); document.write( "\n" ); document.write( "Thankyou so much...
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Algebra.Com's Answer #114685 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
I'll do the first two to get you started\r
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\n" ); document.write( "\n" ); document.write( "a) Domain of \"f%28x%29=%282%29%2F%28x%5E2%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2B1=0\" Set the denominator equal to zero. Remember, you cannot divide by zero.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=-1\" Subtract 1 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%28-1%29\" or \"x=-sqrt%28-1%29\" Take the square root of both sides\r
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\n" ); document.write( "\n" ); document.write( "Since the square root of negative 1 is not a real number, this means that no real number will make the denominator equal to zero. So there are no domain restrictions. This means that the domain of \"f%28x%29=%282%29%2F%28x%5E2%2B1%29\" is all real numbers.\r
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\n" ); document.write( "\n" ); document.write( "So the domain of f(x) in set-builder notation is which is () in interval notation\r
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\n" ); document.write( "\n" ); document.write( "Domain of \"g%28x%29=sqrt%28x-3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x-3%3E=0\" Set the radicand \"x-3\" greater than or equal to zero\r
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\n" ); document.write( "\n" ); document.write( "\"x%3E=3\" Add 3 to both sides\r
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\n" ); document.write( "\n" ); document.write( "So any number greater than or equal to 3 is in the domain of \"g%28x%29=sqrt%28x-3%29\"\r
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\n" ); document.write( "\n" ); document.write( "So the domain of g(x) in set-builder notation is which is [) in interval notation\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29=%282%29%2F%28x%5E2%2B1%29\" Start with the first function\r
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\n" ); document.write( "\n" ); document.write( "\"f%28g%28x%29%29=%282%29%2F%28%28sqrt%28x-3%29%29%5E2%2B1%29\" Plug in \"g%28x%29=sqrt%28x-3%29\". In other words, replace each \"x\" with \"sqrt%28x-3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%28g%28x%29%29=%282%29%2F%28%28x-3%29%2B1%29\" Square \"sqrt%28x-3%29\" to get \"x-3\". Note: the value of \"x-3\" is now positive. This means that \"x%3E=3\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%28g%28x%29%29=%282%29%2F%28x-2%29\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "Domain of f(g(x)):\r
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\n" ); document.write( "\n" ); document.write( "\"x-2=0\" Set the denominator equal to zero. Remember, you cannot divide by zero.\r
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\n" ); document.write( "\n" ); document.write( "\"x=2\" Add 2 to both sides\r
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\n" ); document.write( "\n" ); document.write( "So if \"x=2\", then the whole denominator is zero. So this means that we must exclude the value \"x=2\" from the domain. However, we specified earlier that x must be greater than or equal to 3. Since 2 is not greater than or equal to 3, we don't have to worry about restricting this value (as it has been already done)\r
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\n" ); document.write( "\n" ); document.write( "So the domain of f(g(x)) in set-builder notation is which is [) in interval notation\r
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