document.write( "Question 975181: Suppose f(x)=x^2 over the domain of f, then:
\n" ); document.write( "(A)f^(-1) (x)=√x
\n" ); document.write( "(B)f^(-1) (x)=x/2
\n" ); document.write( "(C)f^(-1) (x)=-x^2
\n" ); document.write( "(D)f^(-1) (x) does not exist
\n" ); document.write( "(E) none of the above
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Algebra.Com's Answer #596998 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "if \"f%28x%29=x%5E2\", then inverse \"f%5E%28-1%29%28x%29\" will be:\r
\n" ); document.write( "\n" ); document.write( "\"f%28x%29=x%5E2\".......recall that \"f%28x%29=y\", so we have\r
\n" ); document.write( "\n" ); document.write( "\"y=x%5E2\"...........swap \"x\" and \"y\"\r
\n" ); document.write( "\n" ); document.write( "\"x=y%5E2\".......solve for \"y\"\r
\n" ); document.write( "\n" ); document.write( "\"y=+sqrt%28x%29\", your inverse \"f%5E%28-1%29%28x%29\"\r
\n" ); document.write( "\n" ); document.write( "\"f%5E%28-1%29%28x%29+=+sqrt%28x%29\"\r
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\n" ); document.write( "\n" ); document.write( "so, your answer is: (A) \"f%5E%28-1%29%28x%29+=+sqrt%28x%29\"
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