document.write( "Question 1181070: If g(x) is real valued function defined on set of real numbers satisfying following three conditions:
\n" ); document.write( "(a) g'(0) = 0 (derivative of g at zero is equal to zero)
\n" ); document.write( "(b) g'' (-1) > 0 (second derivative of g at -1 is more than zero)
\n" ); document.write( "(c) g''(x) < 0 if x lies in (0, 2) where (0, 2) os open interval
\n" ); document.write( "What will be grpah of g(x)?
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Algebra.Com's Answer #810913 by greenestamps(13216)\"\" \"About 
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\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( "\"graph%28400%2C400%2C-4%2C4%2C-50%2C50%2C-x%5E3%29\"

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