week9 - The second derivative test for relative extrema Let...

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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 real-life problems, where one may want to maximize the profit or the revenue, minimize the cost, etc......
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week9 - The second derivative test for relative extrema Let...

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