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Unformatted text preview: Chapter 5 What the Derivative tells us about a function 5.1 A zero of a function is a place where f ( x ) = 0. (a) Find the zeros, local maxima, and minima of the polynomial y = f ( x ) = x 3 3 x (b) Find the local minima and maxima of the polynomial y = f ( x ) = (2 / 3) x 3 3 x 2 + 4 x . (c) A point of inflection is a point at which the second derivative changes sign. Determine whether each of the polynomials given in parts (a) and (b) have an inflection point. Detailed Solution: (a) y = f ( x ) = x 3 3 x = x ( x 2 3). Thus zeros are at x = 0, x = 3. To find critical points, we take the derivative and set it equal to 0: dy dx = f prime ( x ) = 3 x 2 3 = 0 when x = 1 . To determine the type we find d 2 y/dx 2 = f primeprime ( x ) = 3(2 x ). Plugging in we find f primeprime (1) = 6 > so x = 1 is a minimum. Similarly we find that x = 1 is a maximum. (b) To find critical points calculate dy dx = f prime ( x ) = 2 x 2 6 x + 4 = 2( x 2 3 x + 2) = 2( x 1)( x 2) = 0 for x = 2, 1. Also, d 2 y/dx 2 = f primeprime ( x ) = 4 x 6; this is positive for x > 3 / 2 and negative for x < 3 / 2 so x = 2 is a minimum and x = 1 is a maximum. (c) In (a) we found that f primeprime ( x ) = 3(2 x ). Clearly this changes sign when x = 0, so x = 0 is an inflection point. Similarly in (b) x = 3 / 2 is an inflection point. v.2005.1  September 4, 2009 1 Math 102 Problems Chapter 5 5.2 Find the absolute maximum and minimum values on the given interval: (a) y = 2 x 2 on 3 x 3 (b) y = ( x 5) 2 on 0 x 6 (c) y = x 2 x 6 on 1 x 3 (d) y = 1 x + x on 4 x  1 2 . Detailed Solution: (a) Making the first derivative equal to zero: y prime = 4 x = 0. The function has one critical value at x = 0, which lies in the interval [ 3 , 3]. When x < 0, y prime < 0 so y is decreasing. When x > 0, y prime > 0 so y is increasing. Thus y has a minimum value of 0 at x = 0. Evaluating y at the end points of the interval we have y = 18 at x = 3 and y = 18 at x = 3. Therefore, the maximum value of y in [ 3 , 3] is 18 and the minimum value is 0. (b) First we find f prime ( x ), equate it to zero to find the critical points: f prime ( x ) = 2( x 5) = 0 x = 5 y = 0 when x = 5. y primeprime ( x ) = 2 > 0 so the curve is always concave upward. y = 0 is therefore a minimum value. Checking the end points: y (0) = 25 and y (6) = 1. On [0 , 6], y = ( x 5) 2 has a maximum of 25 and a minimum of 0. (c) f prime ( x ) = 2 x 1. Setting f prime to zero gives x = 1 2 , which lies outside of the given interval [1 , 3]. The maximum and minimum values are therefore obtained from the end points. At x = 1, y = 6. At x = 3, y = 0. So on [1 , 3], y = x 2 x 6 has a maximum value of 0 and a minimum value of 6....
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This note was uploaded on 05/19/2011 for the course MATH 102 Math 102 taught by Professor Allard during the Winter '09 term at The University of British Columbia.
 Winter '09
 Allard

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