document.write( "Question 999602: Please check my work for max/min problem.\r
\n" ); document.write( "\n" ); document.write( "Find the max/min of:
\n" ); document.write( "f(x) = 2x^3 - 3x^2 + 6 of f on [-1,1]\r
\n" ); document.write( "\n" ); document.write( "f'(x) = 6x^2 - 6x
\n" ); document.write( "f'(x) = 6x(x-1)
\n" ); document.write( "therefore, x = 0 & x = 1 and including the boundary x = -1\r
\n" ); document.write( "\n" ); document.write( "testing...
\n" ); document.write( "f\"(x) = 12x - 6
\n" ); document.write( "f\"(0) = 12(0) - 6 = -6 < 0, MAX value occurs at x = 0
\n" ); document.write( "f'(1) = 12(1) - 6 = +6 > 0, MIN value occurs at x = 1
\n" ); document.write( "f\"(-1) = 12(-1) - 6 = -18 < 0, MAX value occuras at x = -1\r
\n" ); document.write( "\n" ); document.write( "This confuses me because now I have one MIN value that occurs at x = 1, but this isn't actually the value of the MIN just where it occurs right?
\n" ); document.write( "And then I have two MAX's which one is the real max? because I think I am only dealing with rel max/min so I don't think two would make sense.\r
\n" ); document.write( "\n" ); document.write( "Plese explain
\n" ); document.write( "Thank you\r
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Algebra.Com's Answer #617197 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Yes, \"x=0\" and \"x=1\" are two extrema values.
\n" ); document.write( "\"x=0\" is the relative maximum.
\n" ); document.write( "\"x=1\" is the relative minimum.
\n" ); document.write( "However since you're in an interval, you need to check the endpoints and there is a definite min and max.
\n" ); document.write( "\"f%281%29=2%281%29%5E3-3%281%29%5E2%2B6=2-3%2B6=5\"
\n" ); document.write( "\"f%28-1%29=2%28-1%29%5E3-3%28-1%29%5E2%2B6=-2-3%2B6=1\"
\n" ); document.write( "So in the interval, the minimum occurs at \"x=-1\" and the maximum occurs at \"x=0\".\r
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