MVT and Rolle s Theorem

Transcription

1 AP Calculus CHAPTER 4 WORKSHEET APPLICATIONS OF DIFFERENTIATION MVT and Rolle s Teorem Name Seat # Date UNLESS INDICATED, DO NOT USE YOUR CALCULATOR FOR ANY OF THESE QUESTIONS In problems 1 and, state wy Rolle s Teorem does not apply to te function even toug tere exist a and b suc tat f a f b. 1. f x1 x 1. f x cot x. Determine weter te Mean Value Teorem (MVT) applies to te function f x x x on te interval [ 1, ]. If it applies, find all te value(s) of c guaranteed by te MVT for te indicated interval. x 1 on te interval [, ]. If it x applies, find all te value(s) of c guaranteed by te MVT for te indicated interval. 4. Determine weter te MVT applies to te function f x 5. Consider te grap of te function gx x 1 sown to te rigt. a) On te drawing provided, draw te secant line troug te points ( 1, ) and (, 5). b) Since g is bot continuous and differentiable, te MVT guarantees te existence of a tangent line(s) to te grap parallel to te secant line. Sketc suc line(s) on te drawing. c) Use your sketc from part (b) to visually estimate te x-coordinate at te point of tangency. d) Tat x-coordinate at te point of tangency is te value of c promised by te MVT. Verify your answer to part (c) by using te conclusion of te MVT on te interval [ 1, ] to find c. 6. Given x x, explain wy te ypotesis of te MVT are met on [0, 8] but are not met on [ 1, 8].

2 Te MVT and Rolle s Teorem are not te only existence teorems we ave seen tis year. Te Intermediate Value Teorem (IVT) is also an important existence teorem for Calculus (see below.) Intermediate Value Teorem (IVT): Conditions: let f x be a continuous function on te closed interval [a, b] and let k be any number between f a and f b. Consequence: tere is at least one value c in [a, b] suc tat f c k. Use te appropriate teorem (eiter IVT or MVT or Rolle s) to answer te remaining problems. 7. Te grap to te rigt sows portions of a continuous function f. a) Explain wy f must ave a root on te interval (1, 4). b) Must f ave a orizontal tangent line on te interval (1, 4)? Explain wy or wy not. c) Wat additional condition sould te function f meet to guarantee te existence of a orizontal tangent line on te interval (1, 4)? f(x) For questions 8 troug 1, determine weter te statements must be true, migt be true, or cannot be true. Justify your answers. 8. If f 1 0 and f 5 0, ten tere must be a number c in (1, 5) suc tat f c If f is continuous on [1, 5] and f 1 0 and 5 0 suc tat f c If f is continuous on [1, 5] and f 1 and 5 suc tat f ' c If f is differentiable on [1, 5] and f 1 and f 5 (1, 5) suc tat f ' c 0. f, ten tere must be a number c in (1, 5) f, ten tere must be a number c in (1, 5), ten tere must be a number c in 1. If te grap of a function as tree x-intercepts, ten it must ave at least two points at wic its tangent line is orizontal. Some of te following questions require using te IVT and MVT backwards. Tis means tat a fact is stated and you need to identify wat teorem was used to guarantee tat fact. You migt want to read again te conclusions for te IVT and MVT before attempting tese problems! 1. Given x x x 1 ' p 5 on te interval [0, ], will tere be a value p suc tat 0 < p < and? Justify your answer. If your answer is yes, find p.

3 14. Given gx x x x r 11 on te interval [1, ], will tere be a value r suc tat 1 < r < and g? Justify your answer. If your answer is yes, use your calculator to find r. You may use a calculator for tis problem. 15. Te eigt of an object t seconds after it is dropped from a eigt of 500 meters is t 4.9t 500. a) Find te average velocity of te object during te first tree seconds. Remember: average velocity is equal to cange in position divided by cange in time. b) Sow tat at some time during te first tree seconds of fall te instantaneous velocity must equal te average velocity you found in part (a). Ten, find tat time. 16. Te functions f and g are twice differentiable for all real numbers. Te table below sows values of te functions and teir derivatives at selected values of x. Te function is given by x f xgx. x f(x) g(x) f (x) g (x) a) Explain wy tere must be a value r for 1 < r < 4 suc tat r. b) Explain wy tere must be a value p for 1 < p < 4 suc tat ' p 11. c) Is te function increasing or decreasing wen x =? Justify your answer.

4 AP Calculus CHAPTER 4 WORKSHEET ANSWER KEY APPLICATIONS OF DIFFERENTIATION MVT and Rolle s Teorem 1. Rolle s Teorem does not apply because f x1 x 1 interval ( 1, ). Its derivative does not exist at x = 1.. Rolle s Teorem does not apply because f x cot x interval, 4. Since f x x x 5 4. It is discontinuous at x = π. is not differentiable for all values on te is not continuous for all values on te is a polynomial, it is bot continuous and differentiable on te interval [ 1, ]MVT applies. f f c 1 c 1 4. Since f x x 1 is discontinuous at x = 0, te MVT does not apply on te interval [, ]. x 5. a) See drawing (on solid green.) b) See drawing (on dased orange.) 1 c) x g g1 5 1 d) c c 1, ' x does not exist (tat is, (x) is not differentiable) at x = 0. x Terefore, (x) is continuous in [0, 8] and differentiable in (0, 8) MVT applies. However, (x) 1 6. Since ' x x 1 is continuous in [ 1, 8] but not differentiable in ( 1, 8) MVT does not apply. 7. a) Since f(x) is continuous, te IVT applies. Since tere are values were x 0 were f x 0, tere must be a value c were f c 0. f and values b) No, not necessarily. We could just connect te grap wit a line segment making f(x) continuous but not differentiable. c) f(x) sould be differentiable.

5 8. Migt be true, but if te function is not continuous it could be false. 9. Te IVT guarantees tis is always true. 10. Migt be true, but if te function is not differentiable it could be false. 11. Te MVT or Rolle s Teorem guarantee tis is always true. 1. Migt be true, but if te function is not differentiable it could be false. 1. Since x x x 1 is a polynomial, it is bot continuous and differentiable on te interval [0, ]MVT applies. Using it: 5 ' p. So, yes suc value exists. 0 5 p 1 p. However, since p must be inside (0, ) p 14. Since gx x x x Since: g 1 and g 1 gx is a polynomial, it is continuous on te interval [1, ] IVT applies. 1 will reac all values between 1 and 1, including 11. So, yes suc value exists. 11 r r r r a) Average velocity 14.7 ft/sec 0 b) Since t 4.9t 500 is a polynomial, it is bot continuous and differentiable on te 0 interval [0, ]MVT applies. Using it: ' c. So te average velocity, 0 0, will equal te instantaneous velocity, ' t, at some time c c c 1.5 sec a) Since f and g are continuous functions (because tey are differentiable), so is x f x gx Intermediate Value Teorem applies to x in [1, 4]. Wit 1 f 1 g tere must be a value r for 4 f 4 g4 9 1 < r < 4 suc tat 9 4 r 1 4. b) Since f and g are continuous and differentiable functions, so is x f x g x x in [1, 4]. Wit Mean Value Teorem applies to 1 f 1 g f 4 g So tere must be a value p for 1 < p < 4 suc tat 4 1 ' p ' x f ' x g x f x g' x ' 10 is decreasing at x = because ()<0. c)