document.write( "Question 1131288: Using the intermediate value theorem, determine, if possible, whether the function f has at least one real zero between a and b. f(x)=x^(3)+4x^(2)-6x-13; a=-8, b=-4 \n" ); document.write( "
Algebra.Com's Answer #747915 by MathLover1(20850)\"\" \"About 
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\"f%28x%29=x%5E%283%29%2B4x%5E%282%29-6x-13\"\r
\n" ); document.write( "\n" ); document.write( "\"a=-8\", \"b=-4+\"\r
\n" ); document.write( "\n" ); document.write( "\"f%28-8%29=%28-8%29%5E%283%29%2B4%28-8%29%5E%282%29-6%28-8%29-13\"
\n" ); document.write( "\"f%28-8%29=-221\"}\r
\n" ); document.write( "\n" ); document.write( "\"f%28-4%29=%28-4%29%5E%283%29%2B4%28-4%29%5E%282%29-6%28-4%29-13\"
\n" ); document.write( "\"f%28-4%29=11\"\r
\n" ); document.write( "\n" ); document.write( "Now we know:\r
\n" ); document.write( "\n" ); document.write( "at \"x=-221\", the curve is \"below\" zero
\n" ); document.write( "at \"x=11\", the curve is \"above+\"zero\r
\n" ); document.write( "\n" ); document.write( "And, being a polynomial, the curve will be continuous, so somewhere in between the curve must cross through \"f%28x%29=0\"\r
\n" ); document.write( "\n" ); document.write( "Yes, there is a solution to \"x%5E%283%29%2B4x%5E%282%29-6x-13=+0\" in the interval [\"-8\", \"-4\"]\r
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\n" ); document.write( "\n" ); document.write( "\"+graph%28+600%2C+600%2C+-15%2C+15%2C+-10%2C10%2C+x%5E%283%29%2B4x%5E%282%29-6x-13%29+\" \r
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