document.write( "Question 1181070: If g(x) is real valued function defined on set of real numbers satisfying following three conditions:
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document.write( "(a) g'(0) = 0 (derivative of g at zero is equal to zero)
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document.write( "(b) g'' (-1) > 0 (second derivative of g at -1 is more than zero)
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document.write( "(c) g''(x) < 0 if x lies in (0, 2) where (0, 2) os open interval
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document.write( "What will be grpah of g(x)? \n" );
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Algebra.Com's Answer #810913 by greenestamps(13216) You can put this solution on YOUR website! \n" ); document.write( "You can't tell what the whole graph is going to look like; you can only determine certain features of the graph. \n" ); document.write( "(a) g'(0) = 0 (derivative of g at zero is equal to zero) \n" ); document.write( "The instantaneous slope of the graph is 0 at x=0, which means the tangent to the graph at x=0 is horizontal. This might be a maximum or minimum; or it might be a point of inflection. \n" ); document.write( "(b) g'' (-1) > 0 (second derivative of g at -1 is more than zero) \n" ); document.write( "Second derivative greater than zero means the graph is concave up; so at x=-1 the graph is concave up. \n" ); document.write( "(c) g''(x) < 0 if x lies in (0, 2) where (0, 2) is open interval \n" ); document.write( "Second derivative less than zero means the graph is concave down; so the graph is concave down on the interval (0,2). \n" ); document.write( "With the graph concave up at x=-1, the tangent to the graph horizontal at x=0, and the graph concave down on the interval (0,2), the simplest polynomial function will be a cubic polynomial with an inflection point at x=0. \n" ); document.write( "The function g(x)=-x^3 satisfies all those conditions: \n" ); document.write( "g'(x) = -3x^2; g'(0)=0 \n" ); document.write( "g''(x) = -6x; g''(-1)=6 which is greater than 0; and g''(x) is negative for all positive values of x, including the interval (0,2). \n" ); document.write( "A graph.... \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |