document.write( "Question 1037050: Find the equations on the real number p for which $\"x%5E4+-4p%5E3x%2B12%3E0\"$ for all real numbers x . \n" ); document.write( "

Algebra.Com's Answer #651878 by robertb(5830) $\"\"$ $\"About$

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Find the extreme points of the function $\"f%28x%29+=+x%5E4+-4p%5E3x%2B12\"$ :\r
\n" ); document.write( "\n" ); document.write( "==> f'(x) = $\"4x%5E3-4p%5E3\"$ = 0
\n" ); document.write( "==> $\"x%5E3-p%5E3+=+%28x-p%29%28x%5E2%2Bxp%2Bp%5E2%29+=+0\"$
\n" ); document.write( "==> There is only one critical value of x = p. ( The function $\"y+=+x%5E2%2Bxp%2Bp%5E2\"$ has no real roots in terms of p and so there no critical values coming from this function.)\r
\n" ); document.write( "\n" ); document.write( "Because f\"(p) = $\"12p%5E2\"$ >0, the second derivative test tells that there is a relative (in fact, absolute) minimum at x = p. \r
\n" ); document.write( "\n" ); document.write( "We want $\"f%28p%29+=+p%5E4+-4p%5E3%2Ap%2B12++=+p%5E4+-4p%5E4%2B12+=+12+-+3p%5E4%3E+0\"$ , or \r
\n" ); document.write( "\n" ); document.write( " $\"4+-+p%5E4+=+%282+-+p%5E2%29%282+%2B+p%5E2%29+%3E+0\"$ .
\n" ); document.write( "Since $\"2%2Bp%5E2+%3E+0\"$ always, it follows that $\"2+-+p%5E2+%3E+0\"$ , or $\"2+%3E+p%5E2\"$ .\r
\n" ); document.write( "\n" ); document.write( "The solution set to the last inequality is $\"-sqrt%282%29+%3C+p+%3C+sqrt%282%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "Therefore as long as p is in the open interval ( $\"-sqrt%282%29\"$ , $\"sqrt%282%29\"$ ), $\"x%5E4+-4p%5E3x%2B12%3E0\"$ for all real numbers x .
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