AP Calculus AB Sample Exam Questions Course and Exam Description Effective Fall 2016

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1 P alculus Sample Eam Questions ourse and Eam escription Effective Fall 6 Section I, Part ( graphing calculator may not be used) Multiple hoice Questions lim f( g( )) f(lim g( )) f() 3 7 sin lim Since 7 sin and sin3 are differentiable on an sin3 interval around =, and 7 sin and sin3 then then by L Hopital s Rule, 7 sin 7cos 7 6 lim lim sin3 3 cos3 () 3() 3 f( ) g( h( j( )) so where g( ) sin ; h( ) ln ; and j( ) f '( ) cos ln g'( ) cos ; h'( ) ; and j'( ) and cos ln f '( ) The graph III has a slope of zero at about =.5 and = 3. Graph I shows y(~.5) is zero and y(~3) is zero. So I is the derivative of III. The graph III has a slope of zero at about = and about =.5. The graph III has y values of at about =,.3, and.75 which correspond to the places on graph II where the slope is zero. So graph III is the graph of the derivative of graph II. Graph II is increasing in the interval.5 and 5 6. Graph III is the only possibility for the graph of the derivative of graph II. The slopes of graph III go from being positive to negative to positive. This is reflected in graph I. So graph II is f, graph III is f and graph I is f. student might use different arguments to also reach the same conclusion. Since the local linear approimation to g at = ½ is y=+, g ' and g 3, so g g' 7. Math Through iscovery 6

2 at () v'() t t5t t t t t t t t t 5 ( ) 3 3 t, Since is the cost of shredding pounds of documents (5)=8 is an instantaneous rate of change of the cost per pound is $8 per pound when the weight of the documents is 5 pounds. n 3k 3 3 3n lim is a right n k n n n n n n n n Riemann Sum for a function y 3 from = to =. b a ;, n n. So the correct integral would n n n n n be 3d Making a sketch of f: y 9. then 5 f d f d f d 3 or. 5 f d f d f d d 3d 3 u e ; du e d y substitution the integral becomes sin cos u du u c and then by re substitution the integral becomes sin e c Math Through iscovery 6

3 ..t The growth rate is 5e t so the change in the population from day to day 5 e..t 5e would be e e.. e.t t dt t 5 5 5e 75 The population density miles from the highway is given by ( ) thousands of people representative re sq. mi. The dimensions of the representative rectangle units from the highway is and high. The number of people miles from the highway is ( ) area of rectangle. The total population is found by. 3. ( d ) ( ) ( ) d. dy ysec d dy sec d y ln y tan c, ln tan c c c ln y tan tan y e ut since y we select y e tan Math Through iscovery 6

4 . For a function f to be continuous at a point =a the function must have a function value and the lim f( ) f( a). t =, = the function is not a defined. t = the lim f( ) does not eist. Therefore, the function f is not continuous at three locations: =, +, and +. From the chart we know that f() & g(). Since f and g are inverse functions g '(). Therefore, g '() f '() 5 5. Section I, Part (graphing calculator required for some questions) Multiple hoice Questions For a function f to have a relative minimum the derivative of f must change from negative to positive (same as the function f changing from decreasing to increasing). This occurs at = For a function g to be concave up the second derivative must be positive. Looking at a graph of g. From the graph we can tell that g is negative when is between =.6 and =5. Math Through iscovery 6

5 If the temperature of the water is given as H(t), where t is measured in days since Jan, the H (9) is the instantaneous rate of change of the temperature of the water on day o H'(9).53 / day h ( ) f( ). Since f is a differentiable function on its domain, f() is also differentiable. Since y= is a continuous function, f() or h() is a continuous function on its domain. I. There is no guarantee that between << h is increasing since a differentiable function may increase and decrease over it s interval. h() f() 3 II. so because h is continuous on (,), by the h() f(8) 3 Intermediate Value Theorem there must be a point c in (,) where hc ( ). h() f() 3 III.. The average rate of change between these two h() f() 9 points is so because h is both differentiable on (,) and continuous on (,) there must be a point c in (,) where h'( c) 3 by the Mean Value Theorem.. g() g() h( ) d 3.5 Math Through iscovery 6