document.write( "Question 1108325: Find an interval over which the functio
\n" ); document.write( "F(x)=-(x+5)^2(x-4)^3 is decreasing. \n" ); document.write( "
Algebra.Com's Answer #723344 by Boreal(15235)\"\" \"About 
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-(x+5)^2(x-4)^3
\n" ); document.write( "derivative is 2(x+5)(-1)(x-4)^3+3(x-4)^2*-(x+5)^2
\n" ); document.write( "The minus signs cancel and set equal to 0.
\n" ); document.write( "2(x+5)(x-4)^3+3(x-4)^2(x+5)^2=0
\n" ); document.write( "at x=-5 or x=4, the derivative has a critical point.
\n" ); document.write( "there should be one more.
\n" ); document.write( "2(x+5)(x-4)^3=-3(x-4)^2*2(x+5)^2
\n" ); document.write( "2(x-4)^3=-3(x-4)^2*2(x+5)
\n" ); document.write( "2(x-4)=-3*2(x+5)
\n" ); document.write( "2X-8=-6X-30
\n" ); document.write( "8x=-22
\n" ); document.write( "x=-22/8
\n" ); document.write( "Graphing is a good way to see it.
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C10%2C-100%2C5000%2C-%28x%2B5%29%5E2%28x-4%29%5E3%29\"
\n" ); document.write( "Without the graph, one can look at extreme values. As x becomes more negative, then it is minus the square* minus^3, and that is positive. So from x going from -oo to -5, the function is decreasing.
\n" ); document.write( "As x gets large positive, it is decreasing, so from -22/8 ( a local maximum), to +oo, it is decreasing.\r
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