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1 Name: Date: Block: Semester Assessment Revision 3 Multiple Choice Calculator Active NOTE: The eact numerical value of the correct answer may not always appear among the choices given. When this happens, select from among the choices the number that best approimates the eact numerical value.. Let f be the function given by f() = tan and let g be the function given by g() = 2. At what value of in the interval 0 do the graphs of f and g have parallel tangent lines? (a) 0 (b) (c) (d) 2.94 (e) Let ft () for t 0. For what value of t is t closed interval [a, b]? (a) ab (d) ab 2 b a (b) ab (e) (c) ab f t equal to the average rate of change of f on the 3. The figure above shows a road running in the shape of a parabola from the bottom of a hill at A to point B. At B, it changes to a line and continues to on to C. The equation of the road is 2 R () a, From AtoB b c, From BtoC B is,000 feet from A and 00 feet higher. Since the road is smooth, value of b? R is continuous. What is the (a) 0.2 (b) 0.02 (c) (d) (e)

2 4. The figure above shows the graph of the derivative of a function f. How many points of inflection does f have in the interval shown? (a) None (b) One (c) Two (d) Three (e) Four 5. The amount At of a certain item produced in a factory is given by A(t) = (t 3) 4(t 3) 3 where t is the number of hours of production since the beginning of the workday at 8:00 a.m. At what time is the rate of the production increasing most rapidly? (a) 8:00 am (b) 0:00 am (c) :00 am (d) 2:00 noon (e) :00 pm 6. At how many points on the curve through the origin? (a) One (b) Two (c) Three (d) Four (e) Five y will the line tangent to the curve pass 7. The graph of the derivative of a twice differentiable function is shown below. If f () = 2, which of the following must be true? (a) f (2) < f (2) < f (2) (b) f (2) < f (2) < f (2) (c) f (2) < f (2) < f (2) (d) f (2) < f (2) < f (2) (e) f (2) < f (2) < f (2) 8. The function tan 3 f has a zero in the interval [0,.4]. The derivative at this point is (a) 0.4 (b).042 (c) 3.45 (d) (e) undefined

3 9. Let f be a function that is everywhere differentiable. The value of f the table below f If f () is always increasing, which statement about f () must be true? (a) f () has a relative min at = 0. (b) f () is concave down for all. (c) f () has a point of inflection at (0, f (0)) (d) f () passes through the origin (e) f () is an odd function is given for several values of in 20. The table below gives the values of a differentiable function f. what is the approimate value of f (a) (b) (c) (d) (e) f 4 cannot be approimated from the information given. f () Which graph best represents the position of a particle, s(t), as a function of time, if the particle s velocity and acceleration are both positive? 4? 22. (a) I only (b) II only (c) I and II only (d) I and III only (e) II and III only

4 Revision 3: Free Response Calculator Active 3. a) Relative minimum at positive at. b) Relative maimum at 5 negative at 5. c) because f() because f() f is concave down on( 7, 3) (2,3) (3,5) intervals where the slope of f () is negative. changes from negative to changes from positive to because these are the : answer : justification : answer : justification : (-7,-3) : (2, 3) (3,5) d) The absolute maimum occurs at 7. An absolute minimum must occur at either a critical number or at an endpoint, so the possibilities from this graph are at 7, 5 or 7. f( 5) f( 7) because f is increasing on ( 7, 5) as shown by the graph of the first derivative being positive. f(7) f( 5) because the negative change in f from 5 to is less in magnitude than the positive change in f from to 7. : answer of 7 : identifies 5 are candidates ; and 7 : justifies that f(7) f( 5) 4. a) W ( v) v Wv ( ) / When v 20 mph, the wind chill is decreasing at F/ mph : answer : eplanation : units b) c) W(5) and W(60) 3.05 W60 W / So Wv ( ) when v 23.0 Sub v t 20 5t in for W v 0.6 W20 5t (20 5) t W t t 5 W (3) F/ hr : average rate of change : Wvaverage ' () rate of change : value of v : uses v( t) 20 5t : answer : units

5 5. a) h() f( g()) 6 f(2) 63 h(3) f( g(3)) 6 f(4) Since h(3) 5 h() and h is continuous, by the IVT, there eists a value r, r 3, such that hr ( ) 5 : h() 3 : h(3) 7 : conclusion using the IVT b) h(3) h() Since h is continuous and differentiable, by the MVT, there eists a value c, c 3 such that hc ( ) 5. c) w ( ) f( g( )) g( ) w (3) f( g(3)) g(3) ' w (3) f(4) 26 : generic derivative : correct substitution of values : answer (- if no chain rule) : h(3) h() 3 2: conclusion using the MVT - if no reference to continuity/differentiability 3 g 2 d) First step: then use g 2 gg g : answer y-value on g