document.write( "Question 1060525: Use: f(x) = -2x^2 -5x +3\r
\n" ); document.write( "\n" ); document.write( "State the intervals over which f(x) > 0 \n" ); document.write( "

Algebra.Com's Answer #675488 by math_helper(2461) $\"\"$ $\"About$

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$\"+f%28x%29+=+-2x%5E2+-5x+%2B+3+\"$
\n" ); document.write( "f'(x) = $\"+-4x+-5+\"$
\n" ); document.write( "f''(x) = $\"+-4+\"$ \r
\n" ); document.write( "\n" ); document.write( "Since f''(x) < 0, the function is (everywhere) concave down.\r
\n" ); document.write( "\n" ); document.write( "Setting f'(x) = 0: $\"+-4x-5+=+0+\"$
\n" ); document.write( " $\"+x+=+-5%2F4+\"$ \r
\n" ); document.write( "\n" ); document.write( "This is a critical point, it is a maximum or minimum. In this case, since the function is concave down, it is a maximum.

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\n" ); document.write( "\n" ); document.write( "So we know f(x) > 0 at the maximum, and it will be greater than zero for values between the zero crossings of f(x). To find the zero crossings, set f(x) = 0:\r
\n" ); document.write( "\n" ); document.write( " $\"++-2x%5E2+-+5x+%2B+3+=+0+\"$
\n" ); document.write( " $\"++++2x%5E2+%2B+5x+-+3+=+0+\"$
\n" ); document.write( " $\"++%282x-1%29%28x%2B3%29+=+0+\"$
\n" ); document.write( " $\"++x=1%2F2+\"$ or $\"+x=-3+\"$ \r
\n" ); document.write( "\n" ); document.write( "So $\"+f%28x%29+%3E+0+\"$ on the interval $\"+-3+%3C+x+%3C+1%2F2+\"$ \r
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