MCV4U - Chapter 1 Review

Transcription

1 Ì Ì Name: Class: Date: MCV4U - Capter Review Multiple Coice Identify te coice tat best completes te statement or answers te question.. For te function f(x) x 3 x, determine te average rate of cange of f(x) wit respect to x over te interval x 3. a. 30 c. b. d. 5. For f(x) x 6, determine te average rate of cange of f(x) wit respect to x over te interval x 4. a. 3 c. 3 b. 5 d For te function y f(x), wat does te quotient y x represent? a. te average rate of cange for f(x) over te interval x x x b. te derivative of f at x c. te mean value of f(x) d. te instantaneous rate of cange for f(x) at x A function as te following table of values. x (x) Find te following: (x) x 7 a. 55 c. 54 b. d. undefined 5. Find te following: (x) x 7 a. 55 c. 54 b. d. undefined Knowledge Ï 6. Te it of te sequence, 4, Ô 8,, Ô ÓÔ n Ô a. / c. b. infinity d. 0 Ï 7. Te it of te sequence, 3, 4, 5,, () n Ô n, Ô is ÓÔ Ô a. c. 0 b. a it does not exist d. infinity Page of 5

2 Ì Name: 8. Were is te function f(x) tanx discontinuous? a. Every multiple of / b. Every odd multiple of / c. Every even multiple of / d. Te function is continuous everywere. Tinking Ï 9. Te it of te sequence Ô k, k k, k k k, k k k k, Ô is ÓÔ Ô a. c. 0 b. k d. /k 0. Wat can be said regarding te series S n...? a. Te series is aritmetic because a common difference exists. b. Te series is geometric, because a common ratio exists. c. Te sum of te terms do not approac a finite it. d. All of te above statements are true. f(x ) f(x). For te function y f(x), wat does te quotient represent? a. te mean value of f(x) c. te derivative of f at x b. te slope of a secant line for f(x) d. te derivative of f at. Te instantaneous rate of cange for te grap below is: a. c. x + 5 b. d. 3 Ω 3n 3. Te it of te sequence given by te expression: n n 3, n a. is 3/ c. does not exist b. is 0 d. is 3 Ω log n 4. Te it of te sequence given by te expression: n n 3, n a. is c. does not exist b. is 0 d. is negative (naturalnumbers) (natural numbers)

3 Ì Ì Ì Name: Ω 5. Te it of te sequence given by te expression: n n 3n 5, n a. is c. does not exist b. is d. is negative n 3 (naturalnumbers) Sort Answer. Determine weter te function f is continuous at x = 3: Ï x f(x) Ô3 3, x 3 ÓÔ (x 3), x 3. Determine weter te function f is continuous at x = 5: Ï x 4, x 5 f(x) Ô ÓÔ 3x 8, x 5 3. Evaluate te it, using it laws: (x 5x 3) x 0 4. Evaluate te it, using it laws, sowing any algebra: x sinx cosx tanx 5. Evaluate te it, using it laws, sowing any algebra: x x x 6. Evaluate te it, using it laws, sowing any algebra: x x x 7. A large balloon in te sape of a spere as a volume v(r) 4 3 r 3. As te balloon is being inflated, find te average rate of cange of te balloon s volume wen r =, using first principles. (r is in metres). 8. An approximate model for te distance s metres traveled in t seconds by a skateboard going down a ill is s kt, were k is a constant tat depends on te slope of te ill. If, for one particular ill, te value of k is 0., find te instantaneous rate of cange of distance wit respect to time at 0 seconds using first principles. 9. State Caucy s tree conditions for continuous functions. Ï x ifx 0. Sow tat f(x) is discontinuous at x = 0 if f(x) Ô x if x 0 (x ) if x 0 ÓÔ using Caucy. 3

4 Name:. (Communication, ) Given te grap of f(x), determine f(x). x. Wat is te slope of te tangent of f(x) 3x at x? Knowledge State te rate of cange of te given lines 3. y 3x 4. For y 5 x, find te equation for te tangent slope for any x. 5. Find te instantaneous rate of cange (exact) at x = for te function f(x) x x Evaluate te it using it laws: x 3x x Problem. From first principles, find te derivative of: f(x) x /3 x /3.. Find te derivative for f(x) 9 9 x using first principles. 3. If a stone is trown downward wit a speed of 5 m/s from a cliff tat is 80 m ig, its eigt in metres after t seconds is s(t) 80 5t 4.9t. a) Determine wen te stone its te ground. b) Find te stone's instantaneous rate of cange of distance wen it its te ground. 4. Sow tat te curve y x and te curve y x intersect eac oter at rigt angles. 4

5 Name: 5. Evaluate te tangent of te function f(x) x at te point were x a. 5

6 MCV4U - Capter Review Answer Section MULTIPLE CHOICE. ANS: D PTS: REF: Knowledge and Understanding OBJ: 3. Examining Rates of Cange in Polynomial Models STA: CC.0 TOP: Rates of Cange in Polynomial Function Models NOT: from Nelson MCB4U bank. ANS: C PTS: REF: Knowledge and Understanding OBJ: 3. Examining Rates of Cange in Polynomial Models STA: CC.0 TOP: Rates of Cange in Polynomial Function Models NOT: from Nelson MCB4U bank 3. ANS: A PTS: REF: Knowledge and Understanding OBJ: 3. Examining Rates of Cange in Polynomial Models STA: CC.0 TOP: Rates of Cange in Polynomial Function Models NOT: from Nelson MCB text 4. ANS: B PTS: REF: Knowledge and Understanding OBJ: 3.3 Limits of Polynomial Functions STA: DA.0 TOP: Rates of Cange in Polynomial Function Models 5. ANS: D PTS: REF: Knowledge and Understanding OBJ: 3.3 Limits of Polynomial Functions STA: DA.0 TOP: Rates of Cange in Polynomial Function Models 6. ANS: D PTS: DIF: ** REF: K STA: A..4 TOP: Te Limit of a Sequence NOT: PJK - From DDEK (Analysis 3, Gage), Sec 0.3, Page 343 #b 7. ANS: C PTS: DIF: ** REF: K STA: A..4 TOP: Te Limit of a Sequence NOT: PJK - From DDEK (Analysis 3, Gage), Sec 0.3, Page 343 #c 8. ANS: B PTS: DIF: *** REF: K STA: A.. TOP: Continuity NOT: PJK - From DDEK (Analysis 3, Gage), Sec 0.3, Page 36, a modified form of # 9. ANS: B PTS: DIF: ** REF: K STA: A..4 TOP: Te Limit of a Sequence NOT: PJK - From DDEK (Analysis 3, Gage), Sec 0.3, Page 343, a modified form of #4 0. ANS: D PTS: DIF: *** REF: T STA: A..4 TOP: Te Limit of a Sequence NOT: PJK - From DDEK (Analysis 3, Gage), Sec 0.3, Page 343, a modified form of #5. ANS: C PTS: REF: Knowledge and Understanding OBJ: 3.5 Finding Some Sortcuts - Te Constant and Power Rules STA: DA.0 TOP: Rates of Cange in Polynomial Function Models. ANS: A PTS: DIF: ** REF: K TOP: Rate of Cange NOT: PJK - based on text 3. ANS: B PTS: DIF: ** TOP:.3 Limit of a sequence NOT: PJK 4. ANS: B PTS: DIF: ** TOP:.3 Limit of a sequence NOT: PJK 5. ANS: C PTS: DIF: ** TOP:.3 Limit of a sequence NOT: PJK

7 ÁÁÁÁÁ SHORT ANSWER. ANS: First, determine if te it exists: f(x) x 3 f(x) x 3 f(x) f(x) x 3 x 3 So, te it is at x = 3. Second, determine te continuity. Te it is f(3) = and te it = f(3). Terefore, by Caucy, te function is continuous. PTS: DIF: *** REF: T TOP: Continuity (Caucy) NOT: PJK - A piecewise continuity question.. ANS: A solution and grap are provided: First, determine if te it exists: f(x) 3 x 5 f(x) 3 x 5 f(x) f(x) x 5 x 5 So, te it is 3 at x = 5. Second, determine te continuity. Te it is 3 f(5) = 3 and te it = f(5). Terefore, by Caucy, te function is continuous. PTS: DIF: *** REF: T TOP: Continuity (Caucy) NOT: PJK - A piecewise continuity question. 3. ANS: x 0 x 5 x 0 x 3 x 0 3 x 0 x 0 Ê ˆ x 3 x 0 3 PTS: DIF: ** REF: A TOP: Limit laws NOT: PJK

8 ÁÁÁ ÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁÁÁ 4. ANS: x sinx cosx tanx x cosx sinx cos x sinx x Ê cosx sinx x sinx x ˆ cosx x PTS: DIF: *** REF: A TOP: Limit laws NOT: PJK - Algebraic manipulation needed first to avoid undefined 5. ANS: x x x x (x )(x ) x (x ) x x x x 0 PTS: DIF: *** REF: A TOP: Limit laws NOT: PJK - Algebraic manipulation needed first to avoid undefined. A beginning question from Spivak 6. ANS: x x x x x x Ê ˆÊ x x x x x x x / ˆ PTS: DIF: *** REF: A TOP: Limit laws NOT: PJK - Algebraic manipulation needed first to avoid undefined. A beginning question from Spivak 3

9 7. ANS: v(r) 4 3 (r )3 4 3 r È r 3 3r 3r 3 r ÎÍ È 3r 3r ÎÍ È 3r 3r ÎÍ 4r So, tat wen r =, te volume increases at te rate of 4 m 3 /(m radius) (about.568 m 3 per m radius). PTS: DIF: *** REF: Application OBJ:.5 Introduction to Derivatives STA: A.5 TOP: Introduction to Derivatives NOT: PJK - a made-up question for cange in volume 8. ANS: 0.(0 ) 0.(0) s(t) 0.(00 0 ) 0.(0) 0 0 m/s PTS: REF: Application OBJ: 3. Examining Rates of Cange in Polynomial Models STA: CC.0 CC.03 TOP: Rates of Cange in Polynomial Function Models 9. ANS: A function f is continuous at a if ) f(a) is defined ) f(x) exists x a 3) f(x) f(a) x a PTS: DIF: ** REF: Knowledge and Understanding STA: A.. TOP: Definition of Continuity NOT: PJK - Using te Caucy definition of continuity, one tat is rigorous enoug to be used on almost all functions. From Analysis 3, by DDEK, Page 36. 4

10 0. ANS: () f(0) 0 0 () Te left-and it does not include 0, so, we need to approximate: (x ) (.0000) x 0 x 0 x (0) f(x) fails Caucy s test for continuity at x = 0, since te it does not exist (left and rigt-and its are not equal). Terefore, f is not continuous at x = 0. PTS: REF: Knowledge and Understanding OBJ:.4 Limits and Continuity STA: A.4 TOP: Limits and Continuity NOT: PJK - Transmorgrified from NEL AFIC.. ANS: 3 PTS: REF: Knowledge and Understanding OBJ: 5. Limits and End Beaviour of Rational Functions TOP: Rates of Cange in Rational Function Models. ANS: f(x ) f(x) 3(x ) 3x STA: DA.0 3x 6x 3 3x 6x 3 (6x 3) 6x suc tat wen x= /, te tangent as slope 3 PTS: REF: Knowledge and Understanding OBJ: 3.4 Using Limits to Find Instantaneous Rates of Cange - Te Derivative STA: DA.04 TOP: Rates of Cange in Polynomial Function Models NOT: PJK -- Solution 3. ANS: 3(x ) 3x 3x 3 3x 3 3 PTS: DIF: ** REF: K STA: A..4 TOP: Rate of Cange NOT: (PJK) -- DDEK (Analysis 3, Gage, 966), similar to Q on page

11 ÁÁÁÁÁ 4. ANS: 5(x ) /5x / 5x /0x /5 /5x / 0x /5 / (5x 5 /) 5x PTS: REF: Knowledge and Understanding OBJ: 3.5 Finding Some Sortcuts - Te Constant and Power Rules STA: DA.0 TOP: Rates of Cange in Polynomial Function Models 5. ANS: For tis, we need te it. It would be good to first expand f(x ): f(x ) x (x ) 3 x (x 3 3x 3x 3 ) x x 3 6x 6x 3 x x 3 6x 6x 3 x x 3 Tus, 6x 6x 3 6x 6x 6x PTS: DIF: *** REF: A LOC:.5 TOP:.5 Introduction to derivatives NOT: PJK -- # p 0 6. ANS: x 3x x 3x x (x ) x 3 x x Ê ˆ x x 3() x PTS: DIF: ** TOP:.3 Limits NOT: PJK 6

12 PROBLEM. ANS: Tis is best done by breaking up te sum into two separate its, and ence, derivatives. A solution exists in te parametric equations, and relying on te factorzation of a difference of cubes. If your students were not expected to know te sum/difference of cubes formula, ten I would suppose you would need to supply it, but it gives away te secret, and it may be possible tat students may still find it difficult. 3 Let a x (x ) /3 and b x /3. For te negative explonents, let c (x )./ 3 and d x./ 3. d dx x (x ) /3 x /3 /3 x /3 a b c d (x ) /3 x /3 (a b) (a ab b ) (a ab b ) (c d) (c cd d ) (c cd d ) a 3 b 3 (a ab b ) c 3 d 3 (c cd d ) x x ((x ) /3 (x ) /3 x /3 x /3 ) /3 4x 4/3 4x /3 4x x x(x ) x x(x ) ((x ) /3 (x ) /3 x /3 x /3 ) (x ) x (c cd d ) ((x ) /3 (x ) /3 x /3 x /3 )(x)(x ) ((x ) /3 (x ) /3 x /3 x /3 )(x)(x ) 4x 4/3 (x /3 x /3 )(x ) 4x /3 3x 4/3 A conjugate exists and is (x /3 x /3 ), and nearly looks promising, but repeated applications of tis reveals tat te exponent s numerator will always be a power of, and 3 will never divide any power of evenly (you can prove tis by induction). PTS: DIF: **** REF: T TOP:.5 Derivatives by First Principles NOT: PJK - I couldn't solve tis at first, and website researc revealed tat tis ad umbled many oter teacers also. But easily solved by all teacers using te power rule :-) 7

13 . ANS: f (x) d dx 9 9 x 9 9 x 9 9 x Let a 9 x,andb 9 x. f (x) 9 b 9 a 9 b 9 a ( 9 b 9 a ) 9 b 9 a 9 b 9 a b a, substuting on te numerator: ( 9 b 9 a ) 9 x 9 x ( 9 b 9 a ) 9 x 9 x b a 9 x 9 x, cancelling and substituting on te denominator: ( 9 b 9 a )(b a) ( 9 9 x 9 9 x )( 9 x 9 x ) ( 9 9 x x )( 9 x 0 9 x ) 9 9 x 9 x ) x 9 x ) PTS: DIF: **** REF: T TOP:.5 Derivatives by First Principles NOT: PJK - Someting observed as a problem from te 979 Calculus text (Ontario 3). Verified in Maple. 8

14 3. ANS: a) Use te graping calculator to determine wen s(t) 0. t =.79 s b) Find s'(t). s'(t) s(t ) s(t) ÁÁÁÊ 80 5(t ) 4.9(t ˆ ) ÁÁÁÊ 80 5t 4.9t ˆ 80 5t 5 4.9(t t ) 80 5t 4.9t 5 4.9t 9.8t t 5 9.8t 4.9 (5 9.8t 4.9) (5 9.8t 4.9) s(t) 5 9.8t 5 9.8(.79) 4.34 Te instantaneous rate of cange is down at a rate of 4.34 m/s. PTS: REF: Application OBJ: 3.4 Using Limits to Find Instantaneous Rates of Cange - Te Derivative STA: CC.04 DA.04 TOP: Rates of Cange in Polynomial Function Models 9

15 ÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ ÁÁÁ 4. ANS: Find te point of intersection. y x x x = y = Find te slope of te tangents to te curves at te point (, ). y x At x =, y y 3 x At x =, y Using first principles on y = x, letting x = : ( ) f () Te slope at x = is. Bot tangent slopes at te point were te two functions meet are negative reciprocals of eac oter. Terefore, bot tangent lines intersect at rigt angles (are perpendicular to eac oter). Using first principles: (on y = / x wen x = ) f () Ê ˆ ( ) ( ) ( ) Ê ˆ Ê Ê Ê Te slope at x = is /. ˆ ˆ ˆ Ê ˆ Ê ˆ ( ) PTS: DIF: *** REF: Tinking/Inquiry/PS OBJ: 3.6 Finding Some Sortcuts - Te Sum and Difference Rules STA: DA4.0 TOP: Derivatives by First Principles Rates of Cange in Polynomial Function Models NOT: PJK -- First principles answers are mine. Students found tis a callenging question if first principles were used. 0

16 ÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁ 5. ANS: f(x ) f(x) (x ) / x / (x ) / x / (x ) / x / Ê 4/x 4/ x /x 3/ 4 x 4 x x 4 x(x ) x x ˆ x by substitution, te tangent of te function is a Ê 4 x(x 0) x 0 3 4x 4x 4 x(x ) x x ˆ x PTS: DIF: **** REF: Knowledge and Understanding OBJ: 3.5 Finding Some Sortcuts - Te Constant and Power Rules STA: DA.0 TOP: Rates of Cange in Polynomial Function Models NOT: PJK -- a square root problem