document.write( "Question 236145: Evaluate the function at the given value of the independent variable and simplify. f(x) = 3x-1 \r
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\n" ); document.write( "\n" ); document.write( "I started working this problem but I think I'm doing it the wrong way. I'm not sure if I'm suppose to substitue the 3x-1 in each place of f or x?
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Algebra.Com's Answer #173871 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
your equation is f(x) = (3x-1)\r
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\n" ); document.write( "\n" ); document.write( "They want you to evaluate \"%28f%28x%29-f%281%29%29%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "If so, this is how you do it.\r
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\n" ); document.write( "\n" ); document.write( "Let g(x) = \"%28f%28x%29-f%281%29%29%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "You replace f(x) with (3x-1) because that's what f(x) is equal to.\r
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\n" ); document.write( "\n" ); document.write( "You replace x with 1 in f(x) = 3x-1 to get f(1) = 3*1-1 = 2\r
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\n" ); document.write( "\n" ); document.write( "You are left with:\r
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\n" ); document.write( "\n" ); document.write( "g(x) = \"%28f%28x%29-f%281%29%29%2F%28x-1%29\" = \"%28%283x-1%29+-+2%29%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "This becomes:\r
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\n" ); document.write( "\n" ); document.write( "g(x) = \"%283x-3%29%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "You can factor the numerator to get\r
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\n" ); document.write( "\n" ); document.write( "g(x) = \"%283x-3%29%2F%28x-1%29\" = \"%283%2A%28x-1%29%29%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "This becomes:\r
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\n" ); document.write( "\n" ); document.write( "g(x) = 3\r
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