document.write( "Question 1034949: Please help me with this problem:
\n" ); document.write( "Suppose \"f%28x%29+=+x%5E3\" + 2x and \"f%5E-1\" is the inverse of \"f\". Evaluate \"f%5E-1%283%29\" and \"%28f%5E-1%29\"'\"%283%29\".
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Algebra.Com's Answer #649893 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
\"f%28x%29+=+x%5E3+%2B2x+\" ==> \"f%28f%5E-1%28x%29%29+=+x+=+%28f%5E-1%28x%29%29%5E3+%2B+2f%5E-1%28x%29+\" after plugging in the inverse \"f%5E-1%28x%29\" into the original equation.\r
\n" ); document.write( "\n" ); document.write( "Now differentiate implicitly wrt x. \r
\n" ); document.write( "\n" ); document.write( "==> \r
\n" ); document.write( "\n" ); document.write( "Now \"f%28x%29+=+x%5E3+%2B2x+\" ==> \"df%28x%29%2Fdx+=+3x%5E2+%2B2+%3E+0+\" for all real x hence f(x) increasing ==> f(x) is one-to- DISABLED_event_one= => f(x) has an inverse function. Incidentally, f(1) = 3, and so \"f%5E-1%283%29+=+1\".\r
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\n" ); document.write( "\n" ); document.write( "==> \"1+=++%283%2A%28f%5E-1%283%29%29%5E2+%2B+2%29%28df%5E-1%283%29%2Fdx%29++\" \r
\n" ); document.write( "\n" ); document.write( "==> \"1+=+%283%2A1%5E2+%2B+2%29%28df%5E-1%283%29%2Fdx%29+=+5%28df%5E-1%283%29%2Fdx%29\"\r
\n" ); document.write( "\n" ); document.write( "Therefore, \"df%5E-1%283%29%2Fdx+=+1%2F5\"
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