document.write( "Question 638252: good day please help me to solve this problem\r
\n" ); document.write( "\n" ); document.write( "If i have this function :\r
\n" ); document.write( "\n" ); document.write( "f( x ) =1+ 12x – x^3\r
\n" ); document.write( "\n" ); document.write( "which is the minimun relative?
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Algebra.Com's Answer #402081 by Tatiana_Stebko(1539)\"\" \"About 
You can put this solution on YOUR website!
the derivative f'( x )=(1+ 12x – x^3)'=12-3x^2
\n" ); document.write( "\"12-3x%5E2=0\"
\n" ); document.write( "\"x=2\" or \"x=-2\"
\n" ); document.write( "\"f%282%29=1%2B+12%2A2+%96+2%5E3=17\", \"f%28-2%29=1%2B+12%2A%28-2+%29%96+%28-2%29%5E3=-15\"
\n" ); document.write( "(2, 17) and (-2,-15) are the critical points
\n" ); document.write( "Let use Second derivative test
\n" ); document.write( "the second derivative f''( x )=(12-3x^2)=-6x\r
\n" ); document.write( "\n" ); document.write( "at critical point (2, 17) we find f''(2)=\"-6%2A2=-12%3C0\", so (2, 17) is a relative maximum\r
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\n" ); document.write( "\n" ); document.write( "at critical point (-2,-15) we find f''(-2)=\"-6%2A%28-2%29=12%3E0\", so (-2,-15) is a relative minimum
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