Analyzing f, f, and f Solutions
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1 Analyzing f, f, and f Solutions We have intentionally included more material than can be covered in most Student Study Sessions to account for groups that are able to answer the questions at a faster rate. Use your own judgment, based on the group of students, to determine the order and selection of questions to work in the session. Be sure to include a variety of types of questions (multiple choice, free response, calculator, and non-calculator) in the time allotted. Multiple Choice 1. D (199 BC9 appropriate for AB) 1 Since f( ), f (0) does not eist.. B (197 AB) Determine the interval(s) where f( ) 0. 4 f ( ) f ( ) ( ) 0 0 or f ( ) , 0,. on the intervals. D (1997 AB) Determine the interval(s) where f( ) 0. f ( ) ( )( e ) ( )( e ) f( ) e ( ) 0 e ( )( 1) and 1 f( ) 0 on ( 1, ) 4. C (1998 AB19) A point of inflection occurs when f ( ) changes sign. In this case, f ( ) has a double root at and changes sign only at 1 and C (1998 AB) The function, f ( ), is increasing when f( ) 0. f ( ) 4 ( 1) 0 0 f ( ) 4 0,. is positive on the interval Copyright 014 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at
2 Analyzing f, f, and f 6. E (1988 AB4) The function, y, is concave downward when y is negative. 10 y is negative for. ( ) 7. B (199 BC) The function f is decreasing when f is negative. ( ) f e e e ( ) 0 0 or Thus f ( ) 0,0. on the interval 8. B (1997 BC80 appropriate for AB) Using the derivative function on the calculator, graph f. Calculate the smallest value where f changes from increasing to decreasing or decreasing to increasing. or Using the derivative function on the calculator, graph f ( ). Calculate the smallest value where f ( ) changes sign. 9. E (00 BC8 appropriate for AB) Rolle s Theorem guarantees an etreme value on the interval (0, 4) given that f is continuous and differentiable on the closed interval [0, 4] and f(0) f(4). 10. C (00 BC86 appropriate for AB) Analyze the graph of f ( ) to determine the number of times f ( ) changes from increasing to decreasing or decreasing to increasing on the interval ( 1.8, 1.8) or Use the derivative function on the calculator to graph f ( ) to determine the number of times f ( ) changes signs on the interval ( 1.8, 1.8) 11. C (1997 AB85) f ( ) changes sign from positive to negative only at B (1998 AB89) The graph of y 4 is a parabola that changes from positive to negative at and from negative to positive at. Since g is always negative, f changes sign opposite to the way y 4 does. Thus, f has a relative minimum at and a relative maimum at. Copyright 014 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at
3 Free Response AB4/BC1 (a) y 1 cos 01cos 1cos At, there is not an etreme value since y ' does not change signs there. y 1 y 1 Absolute minimum value at, 1 Absolute maimum value at, 1 No point values available. : Analyzing f, f ', and f " (b) y sin sin 0 0, y has points of inflection at (0, 0) and (, ) since y changes signs at 0 and (c) on the interval from,. Copyright 014 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at
4 AB (a) f ( ) e f ( ) e e Analyzing f, f ', and f " No point values available. e (1 ) 0 e 1 0 and pos. neg f is increasing on 0, since f( ) 0in this interval and f is 1 decreasing on,10 since f( ) 0 on this interval. 1 (b) f = 1 is a relative maimum from e part (a) f (0) 0 10 f (10) 0 e 1 1 Therefore,, is the absolute e maimum and (0, 0) is the absolute minimum. Copyright 014 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at
5 Analyzing f, f ', and f " AB4 (a) Slope at 0 is f '(0) At 0, y y( 0) (b) No. Whether f ( ) changes sign at 0 is not known. The only given values of f ( ) is f (0) 0. (c) g( ) e ( f( ) f( )) 0 g(0) e ( f(0) f(0)) () ( ) 0 y40( 0) y 4 (d) g( ) e ( f( ) f( )) g e f f ( ) ( )( ( ) ( )) ( ( ) ( )) e f f e ( 6 f( ) f( ) f( )) 0 g(0) e [( 6)() ( ) (0)] 9 Since g(0) 0 and g 0, g does have a local maimum at 0 4 1: equation 1: answer 1: eplanation 1: g (0) 1: equation : verify derivative 0/ product or chain rule error < -1> algebra error 1: g(0) 0 and g (0) 1: Answer and reasoning Copyright 014 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at
6 AB4/BC4 (a) h( ) 0 at Local minima at and at Note: Since the number line is no longer accepted for justification, also state that h ( ) changes sign from negative to positive at and. (b) h( ) 1 0 for all 0. Therefore, the graph of h is concave up for all : 1 analysis 1: conclusions < -1> not dealing with discontinuity at 0 1: h ( ) 1: h( ) 0 1: answer Analyzing f, f ', and f " 16 7 (c) h(4) 4 7 y ( 4) (d) The tangent line is below the graph because the graph of h is concave up for 4. 1: tangent line equation 1: answer with reason Copyright 014 National Math + Science Initiative, Dallas, TX. All rights reserved. Visit us online at
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