document.write( "Question 1033806: Please help me with this problem:
\n" ); document.write( "Let f be the function f(x) = $\"xe%5E%28-x%5E%282%29%29\"$ where x is a set of real numbers. Find the critical numbers of f. (It is helpful to note that $\"e%5E%28-x%5E%282%29%29\"$ is nonzero for any value of x.) Find the intervals on which f is increasing and on which f is decreasing. Use the information found to tell where f attains local maximum and minimum values. \n" ); document.write( "

Algebra.Com's Answer #649049 by robertb(5830) $\"\"$ $\"About$

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$\"f%28x%29+=+xe%5E%28-x%5E2%29\"$ ==> f'(x) = $\"%281-2x%5E2%29e%5E%28-x%5E2%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "Let f'(x) be equal to 0.\r
\n" ); document.write( "\n" ); document.write( "==> $\"1-2x%5E2=0\"$ ==> $\"x+=+sqrt%282%29%2F2\"$ or $\"x+=+-sqrt%282%29%2F2\"$ , the critical numbers of f(x).\r
\n" ); document.write( "\n" ); document.write( "As you mentioned $\"e%5E%28-x%5E2%29%3E0\"$ , so no solution comes from this.\r
\n" ); document.write( "\n" ); document.write( "In ( $\"-infinity\"$ , $\"-sqrt%282%29%2F2\"$ ), f'(x) < 0, so f(x) is decreasing there.
\n" ); document.write( "In ( $\"-sqrt%282%29%2F2\"$ , $\"sqrt%282%29%2F2\"$ ), f'(x) > 0, so f(x) is increasing there.
\n" ); document.write( "In ( $\"sqrt%282%29%2F2\"$ , $\"infinity\"$ ), f'(x) < 0, so f(x) is decreasing there.\r
\n" ); document.write( "\n" ); document.write( "Therefore f(x) attains local min at $\"x+=+-sqrt%282%29%2F2\"$ , while it attains a local max at $\"x+=+sqrt%282%29%2F2\"$ . \n" ); document.write( "

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