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Unformatted text preview: The second derivative test for relative extrema Let f be a twice differentiable function and x a critical point, that is f ( x ) = 0 . Then: 1) If f 00 ( x ) > , then x is a relative minimum. 2) If f 00 ( x ) < , then x is a relative Maximum. Example: Find the relative extrema for f ( x ) = x 3 + 3 x 2 + 9 x 2 . Solution: First we find the critical points of f : f ( x ) = d dx ( x 3 + 3 x 2 + 9 x 2) = 3 x 2 + 6 x + 9 f ( x ) = 0 ⇒  3 x 2 + 6 x + 9 = 0 ⇒ x 2 2 x 3 = 0 ⇒ ( x + 1)( x 3) = 0 ⇒ x 1 = 1 and x 2 = 3 are critical points. Then we calculate f 00 ( x ) : f 00 ( x ) = d dx ( f ( x )) = d dx ( 3 x 2 + 6 x + 9) = 6 x + 6 Finally, we find the sign of f 00 ( x ) at the critical points of f to decide about the extrema: f 00 ( x 1 ) = f 00 ( 1) = 6 ( 1) + 6 = 6 + 6 > 0 so x 1 = 1 is a relative minimum f 00 ( x 2 ) = f 00 (3) = 6 3 + 6 = 18 + 6 < so x 2 = 3 is a relative maximum Applications to economics of maxima and minima All what you have learned about finding (relative or absolute) maxima and min ima for functions, plus stuff about concavity, has applications in economics and business. Such applications occur in reallife problems, where one may want to maximize the profit or the revenue, minimize the cost, etc......
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