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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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......
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online