document.write( "Question 1194817: Given the functions
\n" ); document.write( "𝑓(𝑥) = 2𝑥 − 1 , 𝑥 ∈ ]1; 10[ ,
\n" ); document.write( "𝑔(𝑥) = 3𝑥 + 1 , 𝑥 ∈ ]−1; two[ . \r
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Algebra.Com's Answer #827113 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "To find the equation for f∘g(x), plug the entire right side of the \r\n" );
document.write( "equation for g(x), for x in the equation for f(x).\r\n" );
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document.write( "f(x) = 2x-1,   g(x) = 3x+1\r\n" );
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document.write( "So we plug (3x+1) in place of x in 2x-1 and get 2(3x+1)-1 which \r\n" );
document.write( "simplifies to 6x+2-1 or 6x+1, so the equation for f∘g(x) is\r\n" );
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document.write( "f∘g(x) = 6x+1.\r\n" );
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document.write( "Now we must find the domain for f∘g(x). \r\n" );
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document.write( "Any value we substitute for x in f∘g(x) must:\r\n" );
document.write( "(a) be in the domain of g(x).\r\n" );
document.write( "(b) g(x) must produce a value in the domain of f(x)\r\n" );
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document.write( "g(x) has the domain  ]-1; 2[ \r\n" );
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document.write( "So the domain for f∘g(x) must be contained entirely within this \r\n" );
document.write( "interval.\r\n" );
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document.write( "Since g(x) is linear, the endpoints of the range of g(x) is found\r\n" );
document.write( "by substituting the end points of the domain of g(x) for x in the\r\n" );
document.write( "equation for g(x): 3(-1)+2=-1, 3(2)+2=8 so g(x) produces the range\r\n" );
document.write( "]-1; 8[ \r\n" );
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document.write( "But f(x) has the domain  ]1; 10[ , so f(x) cannot be used for \r\n" );
document.write( "]-1; 1],\r\n" );
document.write( "so we must remove  ]-1; 1]  from ]-1; 2[ which leaves ]1; 2[ \r\n" );
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document.write( "So:\r\n" );
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document.write( "f∘g(x) = 6x+1, x ∈ ]1; 2[ \r\n" );
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document.write( "Now you switch the roles of f and g and do part b) yourself.\r\n" );
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document.write( "Edwin
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