document.write( "Question 1110853: Find an interval over which the function f(x)=(x+5)^3(x-4)^2 is decreasing.\r
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Algebra.Com's Answer #725862 by KMST(5328) $\"\"$ $\"About$

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The interval over which the function $\"f%28x%29\"$ is decreasing
\n" ); document.write( "is the interval where its derivative is negative.
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\n" ); document.write( " $\"f%28x%29=%28x%2B5%29%5E3%28x-4%29%5E2\"$ can be considered to be the product of the functions
\n" ); document.write( " $\"u%28x%29=%28x%2B5%29%5E3\"$ and $\"v%28x%29=%28x-4%29%5E2\"$
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\n" ); document.write( "You must have learned that for a product function $\"f=u%2Av\"$ ,
\n" ); document.write( "the derivative $\"%22f+%27+%22\"$ can be calculated
\n" ); document.write( "from the factor functions, $\"u\"$ and $\"v\"$ ,
\n" ); document.write( "and their derivatives, $\"%22u+%27+%22\"$ and $\"%22v+%27+%22\"$ , as
\n" ); document.write( " $\"%22f+%27%22+=%22u+%27+%22%2Av%2B%22v+%27%22%2Au\"$ .
\n" ); document.write( "In this case, $\"%22u+%27%22=3%28x%2B5%29%5E2\"$ and $\"%22v+%27%22=2%28x-4%29\"$ , so
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\n" ); document.write( "The zeros of that derivative are $\"x=-5\"$ , $\"x=4\"$ and $\"x=2%2F5\"$ .
\n" ); document.write( "Because $\"x=-5\"$ has even multiplicity
\n" ); document.write( "(multiplicity is 2 because ( $\"%28x-5%29\"$ is squared).
\n" ); document.write( " $\"%22f+%27%22\"$ does not change sign at $\"x=5\"$ .
\n" ); document.write( "the derivative changes sign at the other two roots,
\n" ); document.write( "and it is obviously positive for $\"x%3E4\"$ ,
\n" ); document.write( "so it is negative only in the interval $\"%22%28+2+%2F+5+%2C+4+%29%22\"$ . \n" ); document.write( "

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