document.write( "Question 243300: At what value(s) of x does f(x)= x^4-8x^2 have a relative minimum?
\n" ); document.write( "(A) 0 and -2 only
\n" ); document.write( "(B) 0 and 2 only
\n" ); document.write( "(C) 0 only
\n" ); document.write( "(D) -2 and 2 only
\n" ); document.write( "(E) -2,0, and 2 \n" ); document.write( "
Algebra.Com's Answer #178236 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
At what value(s) of x does \"f%28x%29=+x%5E4-8x%5E2\" have a relative minimum?
\n" ); document.write( "
\r\n" );
document.write( "I assume you are taking calculus:\r\n" );
document.write( "\r\n" );
document.write( "If a function is continuous its relative maximums or minimums they\r\n" );
document.write( "always occur either at points where the derivative is 0 or undefined.\r\n" );
document.write( "\r\n" );
document.write( "(Note: But just because the derivative is 0 or undefined for a certain value of x, \r\n" );
document.write( "that does not always mean that there is a relative maximum or minimum there.)\r\n" );
document.write( "\r\n" );
document.write( "We find the derivative of f(x)\r\n" );
document.write( "\r\n" );
document.write( "\"%22f%28x%29%22=+x%5E4-8x%5E2\"\r\n" );
document.write( "\"%22f%27%28x%29%22+=+4x%5E3-16x\"\r\n" );
document.write( "\r\n" );
document.write( "Set \"%22f%27%28x%29%22=0\"\r\n" );
document.write( "\r\n" );
document.write( "\"4x%5E3-16x=0\"\r\n" );
document.write( "\r\n" );
document.write( "Factor out \"4x\"\r\n" );
document.write( "\r\n" );
document.write( "\"4x%28x%5E2-4%29=0\"\r\n" );
document.write( "\r\n" );
document.write( "Factor the expression in parentheses:\r\n" );
document.write( "\r\n" );
document.write( "\"4x%28x-2%29%28x%2B2%29=0\"\r\n" );
document.write( "\r\n" );
document.write( "Set each factor = 0\r\n" );
document.write( "\r\n" );
document.write( "\"4x=0\" gives \"x=0\"\r\n" );
document.write( "\r\n" );
document.write( "\"x-2=0\" gives \"x=2\"\r\n" );
document.write( "\r\n" );
document.write( "\"x%2B2%29=0\" gives \"x=-2\"\r\n" );
document.write( "\r\n" );
document.write( "So we have 3 critical values.  Now we can find out\r\n" );
document.write( "\r\n" );
document.write( "whether each of these is (1) a relative maximum, (2) a relative minimum,\r\n" );
document.write( "or (3) a horizontal inflection points.\r\n" );
document.write( "\r\n" );
document.write( "by either of two methods, the first derivative test, or\r\n" );
document.write( "the second derivative test.  The second derivative test\r\n" );
document.write( "is the easier, but it has the drawback that it sometimes\r\n" );
document.write( "fails.  The first derivative test is harder, but it never\r\n" );
document.write( "fails.\r\n" );
document.write( "\r\n" );
document.write( "First derivative test.  The intervals to use are the open\r\n" );
document.write( "intervals bounded by the critical values -2, 0 and 2:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Interval    |  (-oo,-2)  |  (-2,0)  |   (0,2)  | (2,oo)\r\n" );
document.write( "Test value  |     -3     |    -1    |     1    |   3\r\n" );
document.write( "Sign of f'  |      -     |     +    |     -    |   +\r\n" );
document.write( "Conclusion  | decreasing |increasing|decreasing|increasing\r\n" );
document.write( "\r\n" );
document.write( "At x = -2 the function changes from decreasing to increasing;\r\n" );
document.write( "therefore there is a relative minimum at the value x=-2.\r\n" );
document.write( "\r\n" );
document.write( "At x = 0 the function changes from increasing to decreasing;\r\n" );
document.write( "therefore there is a relative maximum at the value x=0.\r\n" );
document.write( "\r\n" );
document.write( "At x = 2 the function changes from decreasing to increasing;\r\n" );
document.write( "therefore there is a relative minimum at the value x=2.\r\n" );
document.write( "\r\n" );
document.write( "So the answer is \"f(x) has relative minimums at x = -2 or +2\r\n" );
document.write( "only, choice (D)\r\n" );
document.write( "------------------------------------------------------------\r\n" );
document.write( "\r\n" );
document.write( "The eaier way is to the second derivative test. We find the\r\n" );
document.write( "second derivative:\r\n" );
document.write( "\r\n" );
document.write( "\"%22f%28x%29%22=+x%5E4-8x%5E2\"\r\n" );
document.write( "\"%22f%27%28x%29%22+=+4x%5E3-16x\"\r\n" );
document.write( "\"%22f%27%27%28x%29%22+=12x%5E2-16\"\r\n" );
document.write( "\r\n" );
document.write( "Substitute the critical values in f''(x)\r\n" );
document.write( "\r\n" );
document.write( "\"%22f%27%27%28-2%29%22=+32\"  It's positive, so there is a relative minimum there. \r\n" );
document.write( "\"%22f%27%27%280%29%22=+-16\"  It's negative, so there is a relative maximum there.\r\n" );
document.write( "\"%22f%27%27%282%29%22=+32\"  It's positive, so there is a relative minimum there.\r\n" );
document.write( "\r\n" );
document.write( "Notice this was an easier method, and gives the same answer.\r\n" );
document.write( "\r\n" );
document.write( "When the second derivative is positive the curve is concave upward, and\r\n" );
document.write( "thus we have a minimum.\r\n" );
document.write( "\r\n" );
document.write( "When the second derivative is negative the curve is concave downward, and\r\n" );
document.write( "thus we have a maximum.\r\n" );
document.write( "\r\n" );
document.write( "But when the second derivative turns out to be 0, the test fails, \r\n" );
document.write( "and we must go to the harder first derivative test.\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( " \n" ); document.write( "
\n" );