Calculus 1000 Midterm Booklet Solutions Fall 2019.pdf - \u00a9Prep101 CALC 1000 Booklet Solutions \u2212 A Solving Trig Equations a cos(x \u2212 y sin 2 2 sin \u2212

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Solving Trig Equations . a) cos(x − y) ) sin 2 2 ) sin( − ) Example 2. cos + sin 2 = 0 cos + 2 sin cos = 0 cos (1 + 2 sin ) = 0 cos = 0 3 = , 2 2 1 + 2 sin = 0 sin = −1 2 7 6 6 11 6 6 =+ = = 2 − = . ) = = ) = ( = = LS=RS 1 ©Prep101 CALC 1000 Booklet Solutions b) ( + )2 = 1 + 2 = ( + )( + ) = 2 + 2 + 2 = 1 + 2 = 1 + 2 LS=RS A1. cos(2 ) = − sin2 cos2 − sin2 = − sin2 cos2 − sin2 + sin2 = 0 cos2 = 0 cos = 0 3 = , 2 2 A2. 2 cos2 − cos − 1 = 0 (cos − 1)(2 cos + 1) = 0 cos = 1 2 cos + 1 = 0 = 0,2 2 cos = −1 cos = 2 3 3 =− = −1 2 4 3 3 =+ = 3. tan = sin 2 [0, ] tan = 2 sin cos sin cos = 2 sin cos 2 ©Prep101 CALC 1000 Booklet Solutions sin = 2 sin cos2 0 = 2 sin cos2 − sin 0 = sin (2 cos2 − 1) 2 cos2 − 1 = 0 sin = 0 = 0, cos2 = cos = = 1 2 1 √2 4 cos = −1 √2 3 4 4 =− = ℎ ℎ ℎ , ′ ℎ 4. sin cos [0, ] tan = sin = sin sin = sin cos sin − sin cos = 0 [0, ] sin (1 − cos ) = 0 sin = 0 = 0, , 2 1 = cos = 0, 2 ... 2 is not in our interval ∴ = 0, 3 ©Prep101 CALC 1000 Booklet Solutions A5. a) (1 − )(1 + ) = 2 = (1 − )(1 + ) = 1 + − − 2 = 1 − 2 = 2 LS=RS b) 2 + 2 = = 2 + 2 LS=(2 ) + (1 − 22 ) = 22 + (1 − 22 ) =22 + − 22 =cosx LS=RS c) 1+2 1+ 2 LS= = = 2 1+2 1+ 2 1+ 1+ 2 2 2 2 2 +2 = 2 2 + 2 2 4 ©Prep101 CALC 1000 Booklet Solutions 1 = 2 1 2 1 2 = ( ) 2 1 =2 LS=RS d) 2 − 2 = 2 2 2 2 = − 2 1 2 2 2 = − 2 2 2 (1− 2 ) = 2 = 2 2 LS=RS e) 22 = LS=22 LS=2 ( 1 2 ) = 2 2 = 1 1 ( ) 5 ©Prep101 CALC 1000 Booklet Solutions = LS=RS B. Graphing Exponential and Logarithmic Functions . = 2+1 ∈ (−∞, ∞) ∈ ⁄ > 0 (0, ∞) =0 = − (0,2) . = 3 ∈ (−∞, ∞) ∈ ⁄ > 0 (0, ∞) = 0 − − (0,1) 6 ©Prep101 CALC 1000 Booklet Solutions . = −2 − 4 ∈ (−∞, ∞) ∈ ⁄ < −4 (−∞, −4) = −4 − = − = (0, −5) . = 3(2 ) − 2 ∈ (−∞, ∞) ∈ ⁄ > −2 (−2, ∞) = −2 − = 2 3 ln ln 2 − = (0,1) Work for how to get the intercepts... − = 0 0 = 3(2 ) − 2 3(2 ) = 2 2 = 2 3 ln 2 = ln 2 3 = 2 3 ln ln 2 7 ©Prep101 CALC 1000 Booklet Solutions = 3(20 ) − 2 = 3 − 2 = 1 − = 0 . = (0.5) = ∈ (−∞, ∞) = ∈ ⁄ > 0 (0, ∞) = 0 − = − = (0,1) . = = 0 − = = ln − = (0,1) VA x=0 − (1,0) − () . = 3 log 2 ( − 1) + 3 ↑ Vertical stretch ↑ right 1 ↑ up 3 (multiply y by 3) 8 ©Prep101 CALC 1000 Booklet Solutions . Example . 3 −8 1 4 = 6 −12 2 Example . (3 2 3 )2 = 9 4 6 Example . 1 3 = ( )2−5 27 3 = (3−3 )2−5 = −3(2 − 5) = −6 + 15 7 = 15 = 15 7 Example . 4−2 = , ℎ . Take the ln of both sides ln 4−2 = ln −2 ln 4 = ln −2 ln 4 = ln = 1 9 ©Prep101 C1. a) CALC 1000 Booklet Solutions (2 2 )(92 4 ) −6 2 b) (−2 −1 2 )2 = = 4 183 5 2 −6 2 −2 4 = = −32 4 4 4 2 1 C2 a). 42−1 = ( )2 16 42−1 = (4−2 )2 2 − 1 = −4 3x C2b) 6 =( 2 = −3 1 = −3 2 2x+8 ) 216 63 = (6−3 )2+8 3 = −3(2 + 8) 3 = −6 − 24 9 = −24 = −24 9 = −8 3 2). − = ln 2 ln − = ln(ln 2) − ln = ln(ln 2) = − ln (ln 2) C2 d). (3x-1)(2x-1)=4(3x-1) 6 2 − 3 − 2 + 1 = 12 − 4 6 2 − 17 + 5 = 0 (2x-5)(3x-1)=0 so x=5/2 or x= 1/3 10 ©Prep101 CALC 1000 Booklet Solutions C3. 34−4 = 34 4x-4=4 4x=8 x=2. Therefore, answer is E) C4. 56 = 53 6x=3 x=1/2 Therefore, answer is D) C5. (2−6 )2 = 23 -12=3x x=-4 D. Cancellation Laws Example . cos(cos−1 ( −1 −1 )) = ∈ [−1,1] 2 2 Example . 1 cos [2 ( )] 3 1 Let ( ) = , ℎ 2 = 2 2 − 1 3 11 ©Prep101 CALC 1000 Booklet Solutions 1 1 3 3 Since ( ) = , ℎ = 1 2 2 9 7 3 9 9 9 4 4 3 3 So, 2 2 − 1 = 2 ( ) − 1 = − = − Example . 4 cos [ (− )] 3 Let (− ) = , = − = and since it was arctanx we know it is in the fourth quadrant since it was negative (p.25 of booklet) Find the other side using Pythagorean theorem and we get 5. Now, we want cos = ℎ 3 = . 5 Example . 1 cos [2 ( )] 3 1 Let ( ) = , 2 = 2 2 − 1 3 1 1 3 3 ℎ From ( ) = , = = would be √8 and then cos = √8 . 3 Using Pythagorean theorem, the other side (In first quadrant, since it was arcsinx of a positive x) √8 3 8 16 9 9 So, 2 = 2 2 − 1=2( ) 2 − 1 = 2 ( ) − 1 = 9 7 9 9 − = 12 ©Prep101 CALC 1000 Booklet Solutions −1 −1 2 2 D1. sin(sin−1 ( )) = ∈ [−1,1] D2. −1 cos −1 1 2 arccos ( ) 2 ∴ 2 −1 cos ( ) = 2 cos = −1 −1 2 ℎ ∴ sin( cos( )) = sin = ℎ 2 = −1 3. sin(tan−1 ( ) = sin 2 √3 2 −1 tan−1 ( ) = 2 ∴ 4ℎ 22 + (−1)2 = 2 tan = −1 2 5 = 2 = √5 ∴ sin(tan−1 ( D4. −1 −1 ) = sin = 2 √5 tan( tan 1) = 1 tan ∈ 1 3 2 3 D5. cos(cos ) = cos ( ) = * cos(cos ) = ∈ [0, ] 13 ©Prep101 CALC 1000 Booklet Solutions 1 D6. sin( cos ) = sin 2 1 cos = 2 cos = 1 2 ℎ cos 1 2 . 1 2 , ℎ 1 . 1 √3 sin( cos ) = sin = = 2 ℎ 2 1 D7. tan( cos cos cos = 1 √2 √2 ) ∴ 1 = 1 √2 ℎ ∴ tan( cos 1 1 √2 ) = tan = = 1 1 D8. cos( sin ) 2 1 sin = 2 1 1 4ℎ sin = 1 2 ℎ Since 1/2 is a positive number, we are in the 1st quadrant 1 cos( sin ) = cos() = 2 √3 2 D9. cos( cos √3/√2 ) = √3 2 √3 2 ∈ [−1,1] 14 ©Prep101 CALC 1000 Booklet Solutions D10. cos −1 (tan ) = cos −1 (1) = 0 4 −3 −3 D11. cos(sin−1 ( ) sin1 ( ) = 5 5 sin = −3 5 ∴ sin−1 ℎ 4ℎ (−3)2 + 2 = 52 2 = 25 − 9 2 = 16 =4 −3 4 5 5 cos(sin−1 ( )) = cos = 3 D12. arctan [ ( )] 4 3 4 ( ) = = 1 −1 = −1 using special triangles 3 arctan [ ( )] = arctan(−1) = 4 ℎ ℎ 4ℎ ( . 25) and = −1 =− . 4 15 ©Prep101 CALC 1000 Booklet Solutions 3 D13. cos [ (− )] 5 3 3 5 5 Let (− ) = , ℎ = − ℎ 4ℎ . Using Pythagorean Theorem, we get the other side is 4. 3 5 ℎ We want to find cos [ (− )] = = 4 = from 5 our diagram. p.47 1. no 2. no 3. yes 4. yes E. Inverse . = = 3 +2 3 +2 + 2 = 3 16 ©Prep101 CALC 1000 Booklet Solutions = 3 − 2 = 3−2 Example 4. = ∴ −1 ( ) = = 3−2 2+1 −1 2 + 1 −1 − = 2 + 1 − 2 = 1 + ( − 2) = 1 + = 1+ −2 ∴ −1 ( ) = 1+ −2 Example 5. = √1 − 2 , 0 ≤ ≤ 1 = √1 − 2 square both sides 2 = 1 − 2 2 = − 2 + 1 = √1 − 2 , 0 ≤ ≤ 1 ∴ −1 ( )=√1 − 2 , 0 ≤ ≤ 1 17 ©Prep101 CALC 1000 Booklet Solutions E1. = 3 +5 − 27 = 3 +5 − 27 + 27 = 3 +5 + 27 = +5 3 + 27 ( ) = +5 3 + 27 + 5 = ( ) 3 + 27 = ( )−5 3 ∴ −1 ( )= ( +27 3 )−5 E2. = (3 + 2) = (3 + 2) = (3+2) = 3 + 2 3 = − 2 − 2 = 3 −2 −1 ( )= E3. 3 = 2+5 = 2+5 ln = ln 2+5 ln = (2 + 5) ln ln − 5 = 2 ∴ −1 ( ) = ln − 5 2 18 ©Prep101 E4. = = CALC 1000 Booklet Solutions 1+ 1+ + = = − = (1 − ) = 1− ∴ −1 ( ) = 1− E5. = ln( − 2) = ln( − 2) = ln(−2) = − 2 = + 2 ∴ −1 ( ) = + 2 F. . log 4 ( + 1) + log 4 ( − 2) = 1 log 4 ( + 1)( − 2) = 1 41 = ( + 1)( − 2) 4 = 2 − 2 + − 2 0 = 2 − − 6 0 = ( − 3)( + 2) 19 ©Prep101 = 3, −2 CALC 1000 Booklet Solutions ′ log . ∴=3 . 2 = 8 ln 2 = ln 8 2 ln = ln 8 ln 8 ln 23 3 ln 2 = = = 2 2 2 . log2 3+log2 27 log2 27 log 2 81 log 3 81 = log 2 27 log 3 27 4 = 3 ℎ ∴ ℎ ). . log2 3+log2 81 log2 81 = log2 243 log2 81 = log3 243 log3 81 = 5 4 ℎ ℎ , 20 ©Prep101 CALC 1000 Booklet Solutions F1. 3+5 = 6 ln 3+5 = ln 6 (3 + 5) ln = ln 6 3 ln 6 − 5 = 3 3 = ln 6 − 5 3 ln( 2 + ) = 1 + ln F2. ln( 2 + ) − ln = 1 2 + ln ( )=1 . 2 + = 1 = 2 + 2 + − = 0 2 + (1 − ) = 0 [ + (1 − )] = 0 =0 +1− =0 =−1 ≠ 0 ln 0 F3. log 8 2 + log 2 8 = 1 10 3 3 =3 1 log2 8 ∴ =−1 1 + log 2 8 = + 3 3 21 ©Prep101 CALC 1000 Booklet Solutions F4. log 2 ( + 2) − log 2 (2 + 1) = 3 log 2 ( 23 = 8= +2 )=3 2 + 1 +2 2 + 1 +2 2 + 1 16 + 8 = + 2 15 = 2 − 8 15 = −6 = F5. −6 −2 = 15 5 −2 ) + 1) 5 −4 5 1 = log 2 ( + ) = log 2 ( ) 5 5 5 ℎ log 2 (2 ( ln 4+ln 2 = ln 8 = 8 ∴= −2 5 ∴ ℎ ). F6. 3 log 4 2 + log 4 8 = log 4 23 + log 4 8 = log 4 (8 × 8) = log 4 64 = 3 ∴ ℎ 64 F7. log 4 64 − log 4 16 = log 4 ( ) = log 4 (4) = 1 ∴ ℎ ). 16 F8. log2 25 log2 125 = log5 25 log5 125 = 2 3 ℎ ∴ ℎ ). 22 ©Prep101 log9 3−log2 8 F9. log4 16 1 = 2 −3 log 9 3 = 2 1 =2 −5 = CALC 1000 Booklet Solutions 2 − 6 2 9 = 3 1 32 = 31 2 × 2 −5 = = 4 F10.log 25 125 + log 6 36 + log 4 8 = 1 2 = log 125 log 25 log5 125 log5 25 +2+ +2+ 3 3 2 2 = +2+ log 8 log 4 log2 8 log2 4 =5 F11. = log2 +log2 log2 2 +log2 2 = log2 +log2 2 log2 +2log2 = log2 +log2 2(log2 +log2 ) 1 2 F12. log4 8 log27 9 = log 8 log 4 log 9 log 27 = log2 8 log2 4 log3 9 log3 27 = 3 2 2 3 3 3 9 2 2 4 = × = 23 ©Prep101 CALC 1000 Booklet Solutions *F13. Solve: 3 + 3 ( + 2) = 3 ( − 20) 3 (( + 2) = 3 ( − 20) ( + 2) = − 20 2 + + 20 = 0 (x+5)(x-4)=0 X= -5,4 . Example 1. ) lim ( ) = 3 →−4− lim ( ) = 0 →−4 + ∴ lim ( ) = →−4 ) lim ( ) = −4 →−2 ) lim () = 5 →4+ ) lim ( ) = 5 →4− ) lim ( ) = 5 →4 ) (4) = 5 ) lim ( ) = →0 lim− ( ) = 0 →0 lim+ ( ) = 1 →0 ℎ) (0) = 1 . Example 2. lim ( 2 + 3 ) = 12 + 3(1) = 4 →1 24 ©Prep101 CALC 1000 Booklet Solutions 2 −4 Example 3. lim →2 5 Example 5. lim = 22 −4 5(2) 2 +3+2 +1 →−1 = 0 =0 10 = lim (+1)(+2) (+1) →−1 = lim ( + 2) = −1 + 2 = 1 →−1 (−2)(√−1+1) →2 (√−1−1)(√−1+1) Example 6. lim (−2)(√−1+1) −1−1 →2 = lim (−2)(√−1+1) (−2) →2 = lim = lim(√ − 1 + 1) →2 = √2 − 1 + 1 = 2 2 −5+6 →2 |−2| Example 7. lim lim+ 2 −5+6 & (−2) →2 = lim+ (−2)(−3) →2 (−2) lim− →2 & =2−3 2 −5+6 −(−2) lim− (−2)(−3) −(−2) →2 = = −1 2−3 −1 =1 2 −5+6 →2 |−2| ∴ lim = ℎ = Example 8. lim 5 →0 2 = lim →0 55/5 22/2 25 ©Prep101 CALC 1000 Booklet Solutions 5 5⁄5 = lim 2 →0 2/2 5 lim 5⁄5 = →0 2 lim 2/2 →0 5 = since lim 2 →0 H1. lim 3ℎ2 +ℎ ℎ→0 ℎ+ℎ3 =1 = lim ℎ→0 (. 56 ) ℎ(3ℎ+1) ℎ(1+ℎ2 ) = lim ℎ→0 3ℎ+1 1+ℎ2 = 3(0)+1 1+02 1 = =1 1 √ − 3 √3 − 3 0 = = =0 →3 9 − 9−3 6 2. lim 3. lim 2( 2 −9) −3 →3 4. lim+ −4 →4 1+2 5. lim 3ℎ+ℎ2 ℎ→0 ℎ+ℎ3 = lim 6. lim− →0 (−3) →3 = 4−4 = lim ℎ→0 = lim →0 = lim 2( + 3) = 2(3 + 3) = 12 →3 0 1+2(4) − || 2(−3)(+3) = =0 9 ℎ(3+ℎ) ℎ(1+ℎ2 ) = lim ℎ→0 3+ℎ 1+ℎ2 = lim−(−1) = −1 →0 = 3+0 1+0 =3 ∴ ). ∴ ). ∴ |− | = −() 26 ©Prep101 7. lim CALC 1000 Booklet Solutions 2 −16 →4 −4 8. lim →1 (−4)(+4) = lim (−4) →4 2 −1 = lim ( + 4) = 4 + 4 = 8 →4 (−1)(+1) 2 −2+1 = lim (−1)(−1) = lim (+1) →1 (−1) →1 = 1+1 1−1 2 = =∞ 0 ∴ ℎ ). 9. lim− −2 →2 +2 1 H10. lim ( + →0 1 =lim ( + →0 1 2−2 = 2+2 1 2 − −1+1 →0 (−1) 11. lim ( 4 ) ) = lim ( (−1) = lim ( 0 = =0 1(−1) →0 ) = lim ( + )= →0 (−1) →0 √1+−1 √1++1 ) √1++1) )( 1 ) (−1) 1 0−1 = −1 = lim →0 (√1++1) 1+−1 (√1 + + 1) = lim (√1 + + 1) = √1 + 0 + 1 = 2 →0 →0 = lim 12. lim ( −1 →1 √−1 )( √+1 ) √+1 (−1)(√+1) (−1) →1 = lim = lim (√ + 1) = √1 + 1 = 2 →1 27 ©Prep101 CALC 1000 Booklet Solutions | − 2| − |4 − 2| →0 13. lim − 2 − 2 ≥ 0 ≥ 2 | − 2| = { } −( − 2) − 2 < 0 < 2 |4 − 2| = { 4 − 2 4 − 2 ≥ 0 ≥ 1 2 −(4 − 2) 4 − 2 < 0 < 1} 2 Since x→ 0, | − 2| = −( − 2) |4 − 2| = −(4 − 2) | − 2| − |4 − 2| →0 lim −(−2)−(−1)(4−2) = lim →0 = lim −+2+1(4−2) →0 = lim 3 →0 =lim 3=3 →0 H14. | − 3| = − 3 − 3 ≥ 0 ≥ 3 | − 3| = −( − 3) − 3 < 0 Do both sides... lim →3 |−3| 2 −9 & (−3) = lim+ (−3)(+3) →3 −(−3) lim− (−3)(+3) →3 = lim− −1 →3 +3 = −1 6 28 ©Prep101 lim+ CALC 1000 Booklet Solutions 1 = →3 +3 1 6 | − 3| = ℎ →3 2 − 9 ∴ lim lim H15. = lim |−6|−|+6| | − 6| = − 6 − 6 ≥ 0 ≥ 6 →0 −(−6)−(+6) | − 6| = −( − 6) < 6 →0 = lim −+6−−6 = lim | + 6| = + 6 + 6 ≥ 0 ≥ −6 →0 −2 | + 6| = −( + 6) < −6 →0 Since x→ 0, | − 6| = −( − 2) = −2 | + 6| = + 6 H16. lim (√4+ℎ−2) (√4+ℎ+2) ℎ ℎ→0 = lim 1 ℎ→0 (√4+ℎ+2) (√4+ℎ+2) = 1 √4+2 = = lim (4+ℎ−4) ℎ→0 ℎ(√4+ℎ+2) = lim ℎ = ℎ→0 ℎ (√4+ℎ+2) 1 4 3 H17. lim 4 →0 = lim →0 33/3 44/4 3 3⁄3 = lim 4 →0 4/4 29 ©Prep101 CALC 1000 Booklet Solutions 3 lim 3⁄3 = →0 4 lim 4/4 →0 3 = since lim 4 →0 =1 (. 62 ) I. Example. 2 − 4 = 0 ( − 2)( + 2) = 0 = 2, −2 = 2, −2 Example. (1) = 12 − 4 = −3 lim ( ) = lim ( 2 − 4) = 12 − 4 = −3 →1− →1 lim ( ) = lim (2 − 4) = 2(1) − 4 = −2 →1+ →1 ∴ lim ( ) = →1 ∴ ( ) . = 1 ℎ ℎ Example. = 3 . ℎ ∴ lim ( ) = lim () (32 − 3) = 9 + 3 2 (3) → 3− → 3+ (6) = 9 + 9 2 0 = 9 2 − 6 + 9 divide by 3... 30 ©Prep101 CALC 1000 Booklet Solutions 0 = 3 2 − 2 + 3 = −±√2 −4 2 = = 3 = −2 = 3 2±√4−4(3)(3) 2(3) = 2±√4−36 6 = 2±√−32 6 no solution in the real numbers I1. ( ) = 2 lim ( ) = lim (3 + 1) = 3(2) + 1 = 7 →2− →2 lim+ ( ) = lim ( 2 − 1) = 22 − 1 = 3 →2 →2 ∴ lim− ( ) ≠ lim () →2 →2+ ∴ ( ) . = 2 I2. lim 2 +1 →−1 +1 = (−1)2 +1 −1+1 22 − 4 0 3. (2) = = 2−2 0 2 = =∞ 0 If you factor, the (x-2) cancels and you get x+2...meaning there is a hole in y=x+2 at x=2 4. 2 − 4 = 0 (-2)(+2)=0 31 ©Prep101 CALC 1000 Booklet Solutions = 2, −2 = ±2 5. (1) = 12 − 1 = 0 lim− ( ) = lim ( 2 − 1) = 12 − 1 = 0 →1 →1 lim ( ) = lim (3 − 1) = 3(1) − 1 = 2 →1+ →1 ∴ lim ( ) = →1 ∴ = 1 since left and right hand limits are not equal 6. (−1) = (−1)2 + 4 = 5 lim ( ) = lim ( 2 + 4) = (−1)2 + 4 = 5 →−1+ →−1 lim ( ) = lim ( + 6) = −1 + 6 = 5 →−1− →−1 ∴ lim ( ) = 5 ( ) = −1 →−1 7. . ∴ ℎ ℎ at x=2 (2) + 3 = 2 − 4 2 + 3 = −2 2 = −5 = −5 2 I8. ℎ ℎ , ℎ ( ) . = 1 (1) = 1 = 0 lim ( ) = lim cos −1 ( ) = cos −1 (1) = 0 cos 0 = 1 →1− →1 32 ©Prep101 CALC 1000 Booklet Solutions lim ( ) = lim (ln ) = 1 = 0 →1+ →1 ∴ (1) = lim+ ( ) = lim− ( ) →1 →1 ∴ ( ) . = 1 9. . ∴ ℎ ℎ at x=2 (2) + 2 = (2)2 2 + 2 = 4 2 = 2 =1 I10. . ∴ ℎ ℎ at x=4 (4) − 5 = √4 4 − 5 = 2 2 = 5 = 11. 5 2 lim ( ) = lim+ ( ) = (2) ( ) . = 2 →2− →2 5(2)2 + (2) = 3(2)3 − 5 20 + 2 = 24 − 5 7 = 24 − 20 7 = 4 = 4 7 . 33 ©Prep101 CALC 1000 Booklet Solutions Example 2. lim →∞ 2 2 −+2 2 (2−1+ 22 ) 3 1 →∞ 2 (5+−2 ) = lim 5 2 +3−1 (2−1+ 22 ) 3 1 →∞ (5+−2 ) = lim = 2 5 √16 2 +−4 Example 3. lim −4 →−∞ √16 2 +−4 −4 = lim →−∞ −√ = lim →−∞ 16 2 +−4 2 4 1− 1 4 4 2 −√16 + − = lim →−∞ =− 1− √16 + 0 + 0 1−0 = −4 J1. lim 3 2 −4+6 +5 →∞ J2. lim 2 2 − →∞ 2 +1 3. lim →∞ 4. lim 2 2 +1 →∞ = lim 2 (3−4+ 62 ) →∞ = lim 1 5 2( + 2) 2 (2−1) 1 →∞ 2 (1+2 ) = lim 2 (1) 1 →∞ 2 (1+2 ) 3 2 +5 6 2 +2+1 = lim (3−4+ 62 ) 1 5 ( + 2) = lim = lim →∞ (2−1) 1 →∞ (1+2 ) = lim 1 1 →∞ (1+2 ) 2 (3+5) 2 1 →∞ 2 (6++2 ) = = = lim 2−0 1+0 1 1+0 = 3−0+0 0+0 3 = =∞ 0 =2 =1 (3+5) 2 1 →∞ (6++2 ) = 3+0 6+0+0 = 1 2 34 ©Prep101 CALC 1000 Booklet Solutions J5. √9 2 +−3 −4 = lim →−∞ −√ = lim →−∞ 9 2 +−3 2 4 1− 1 3 4 2 −√9 + − = lim →−∞ 1− = −√9 + 0 + 0 1−0 = −3 J6. lim 3 −2 →−∞ 3−4 3 −2 →−∞ 3−4 = lim = lim →−∞ 2 − 3− 2 4 = ∞2 − 0 3−0 =∞ 35 ©Prep101 CALC 1000 Booklet Solutions 1 1 7. lim sin ( ) = − 1 ≤ sin( ) ≤ 1 →0 8. lim →∞ 2 4 +3 2 +2 4 +2+1 3 2 + ) 2 3 2 1 →∞ 4 (1+3 +4 ) = lim 4 (2+ 3 2 + ) 2 3 2 1 →∞ (1+3 +4 ) = lim (2+ 2 = =2 1 K. Vertical and Horizontal Asymptotes Example 1. ( ) = a) +2 2 −9 2 − 9 = 0 ( − 3)( + 3) = 0 lim +2 →∞ 2 −9 1 2 + 2 9 →∞ 1− 2 = lim = 3, −3 = 0+0 1−0 0 = =0 1 ∴=0 Example2. ( ) = 2 −−12 ( ) = (−4)(+3) (−4) lim ( + 3) = ∞ →∞ K1. ( ) = −3 2 −5−6 −4 = + 3 (cancel) ∴ −3 −6)(+1) =( 36 ©Prep101 CALC 1000 Booklet Solutions = 6, −1 lim 1 3 − 2 5 6 →∞ 1−− 2 −3 = lim →∞ 2 −5−6 = 0−0 1−0−0 =0 =0 K2. ( ) = 4− 2+3 2 + 3 = 0 =− lim 4− →∞ 2+3 3 2 4 −1 3 →∞ 2+ = lim = 0−1 2+0 =− 1 2 ∴=− 1 2 K3. 1 − ln > 0 1 > ln ′ ( ) = < − −1 1−ln 0 -1 undefined 1 − ↘ decreasing on (0, ) ′ ( ) = −1 1 −1 ( )=0 1−ln = 0 = 0 ′ () = 0 37 ©Prep101 CALC 1000 Booklet Solutions = 0, = L. Squeeze Theorem . 1 lim 2 cos ( ) →0 1 − 1 ≤ cos( ) ≤ 1 1 2 (−1) ≤ 2 cos( ) ≤ 1( 2 ) ℎ, lim − 2 = 0 & lim 2 = 0 →0 →0 ↑ ↑ → 0 → 0 → 0 → 0 1 ∴ lim 2 cos ( ) = 0 →0 . lim →∞ 2−cos +3 − 1 ≤ cos ≤ 1 − 1 1 ≥ − cos ≥ −1 ∴ −1 ≤ − cos ≤ 1 Add 2 to each component −1 + 2 ≤ 2 − cos ≤ 1 + 2 ∴ 1 ≤ 2 − cos ≤ 3 Divide by + 3 → ∞, + 3 > 0 38 ©Prep101 CALC 1000 Booklet Solutions 1 ≤ +3 lim 1 2−cos +3 ≤ 3 +3 = 0 = lim →∞ +3 3 →∞ +3 ∴ ℎ lim →∞ 2−cos +3 =0 1 L1. lim 2 sin() →0 1 −1 ≤ sin( ) ≤ 1 1 −1 ≤ sin() ≤ 1 1 2 lim 2 →0 1 ≤ sin() ≤ 1 ≤ 2 sin() ≤ 2 =0 & lim 2 = 0 →0 ∴ ℎ, L2. lim →∞ 1 lim 2 sin() = 0 →0 cos2 (2) 3−2 −1 ≤ cos(2 ) ≤ 1 ∴think about squaring this graph, it would only have values from 0 to 1 39 ©Prep101 CALC 1000 Booklet Solutions divide by 3 − 2 → ∞, 3 − 2 < 0 0 ∴ 3−2 cos2 (2) ≥ 3−2 ≥ 1 3−2 (inequalities switch because we’re dividing by a negative value) ∴ 1 3−2 ≤ Now, lim cos2 (2) 3−2 1 →∞ 3−2 ≤0 =0 ∴ ℎ L3. lim lim 0 = 0 →∞ lim →∞ cos2 (2) 3−2 =0 5 2 −sin(3) →∞ 2 +10 −1 ≤ sin(3 ) ≤ 1 since the sine curve oscillates between -1 and 1. multiply by -1 1 ≥ − sin(3 ) ≥ −1 same as −1 ≤ − sin(3 ) ≤ 1 add 5 2 ℎ 5 2 − 1 ≤ 5 2 − sin(3 ) ≤ 5 2 + 1 5 2 −1 2 +10 ≤ 5 2 −sin(3) 2 +10 ≤ 5 2 +1 2 +10 40 ©Prep101 CALC 1000 Booklet Solutions lim 5 2 −1 →−∞ 2 +10 = lim 5−0 1+0 5 2 +1 →−∞ 2 +10 lim →−∞ 1 = lim 2 (5− 2 ) 10 →−∞ 2 (1+ 2 ) = =5 =5 5 2 −sin(3) 2 +10 = 5 by the Squeeze Theorem . Example 2. ( ) = 2 3 − 3 2 − 1 First, always state that f(x) is continuous on [a,b] or they take off marks:( (1) = 2 − 3 − 1 = −2 < 0 (3) = 2(3)3 − 3(3)2 − 1 = 54 − 27 − 1 > 0 (1) < () < (3) =3>0 ∴ ℎ (1,3)ℎ ℎ () = 0 c 2 3 − 3 2 − 1 = 0 41 ©Prep101 CALC 1000 Booklet Solutions M1. ( ) = 4 + 3 − − 1 First, always state that f(x) is continuous on [a,b] or they take off marks:( (0) = 0 + 0 − 0 − 1 = −1 < 0 (2) = 16 + 8 − 2 − 1 = 24 − 3 = 21 > 0 (0) < () < (2) ∴ ℎ (0,2)ℎ ℎ () = 0 c 4 + 3 − − 1 = 0 2. ( ) = cos they take off marks:( [0, ]First, always state that f(x) is continuous on [a,b] or (0) = cos 0 = 1 > 0 () = cos = −1 < 0 () < () < (0) ∴ ℎ (0, )ℎ ℎ () = 0 , cos = 0 M3. ( ) = sin2 − 3 + 1 (0, 3 2 ) (0) = (sin 0)2 − 03 + 1 = 1 > 0 42 ©Prep101 CALC 1000 Booklet Solutions 3 3 3 3 27 3 2 2 ( ) = sin ( ) − ( ) + 1 = (−1) − ( )+1 2 2 2 8 = ℎ () = sin2 − 3 + 1 = 0 − 3 + 1 = 1 − (3.14)3 < 0 () < () < (0) ∴ ℎ (0, 3 2 ) ℎ ℎ () = 0 ℎ sin2 − 3 + 1 = 0 N. Definition of the Derivative . ( ) − () → − ′ () = lim ′ (3) = lim →3 ()−(3) −3 = lim 3 −33 →3 −3 = lim 3 −27 →3 −3 == 3 2 = lim = 3(3)2 = 27 →3 1 *here you can factor as a difference of cubes, or use L'Hospitals Rule ( − 3)( 2 + 3 + 9) 3 − 27 … lim = lim = lim ( 2 + 3 + 9) = →3 →3 →3 −3 −3 = 32 + 3(3) + 9 = 27 43 ©Prep101 CALC 1000 Booklet Solutions . ( ) = √ + 1 ′ ( ) = lim (+ℎ)−() ℎ ℎ→0 = lim (√+ℎ+1−√+1) (√+ℎ+1+√+1) ℎ ℎ→0 (√+ℎ+1+√+1) +ℎ+1−(+1) = lim ℎ→0 ℎ(√+ℎ+1+√+1) +ℎ+1−−1 = lim ℎ→0 ℎ(√+ℎ+1+√+1) ℎ = lim ℎ→0 ℎ(√+ℎ+1+√+1) 1 = lim ℎ→0 (√ = = + ℎ + 1 + √ + 1) 1 √+1+√+1 1 2√+1 . f(x)=sinx f(x+h)=sin(x+h) ′ ( ) = lim (+ℎ)−() ℎ ℎ→0 Now, from identities sin(x+h)=sinxcosh + cosxsinh ′ ( ) = lim ℎ+ℎ− ℎ ℎ→0 =lim ℎ−+ℎ ℎ→0 = lim ℎ→0 ℎ (ℎ−1)+ℎ ℎ rearrange common factor of sinx from first two terms split up limits 44 ©Prep101 = lim ( ℎ→0 CALC 1000 Booklet Solutions ℎ−1 ℎ ) + lim ( ℎ→0 ℎ ℎ ) =sinx (0)+cosx(1) from limits you've memorized! (I hope!!) p.56 booklet =cosx 45 ©Prep101 CALC 1000 Booklet Solutions 46 ©Prep101 CALC 1000 Booklet Solutions Example 4. () = ( + 2)2 = 3. ( + ℎ) − () ℎ→0 ℎ ′ () = lim (3 + ℎ) − (3) ℎ→0 ℎ ′ (3) = lim (3 + ℎ + 2)2 − (3 + 2)2 ℎ→0 ℎ ′ (3) = lim (5 + ℎ)2 − 25 ℎ→0 ℎ ′ (3) = lim ′ (3) 25 + 10ℎ + ℎ2 − 25 = lim ℎ→0 ℎ ′ (3) ℎ2 + 10ℎ = lim ℎ→0 ℎ ℎ(ℎ + 10) ℎ→0 ℎ ′ (3) = lim m=10 () = ( + 2)2 = 3. Subst. X=3 into f(x) to find y (3) = (3 + 2)2 = 25 Point (3,25) − 25 = 10( − 3) − 25 = 10 − 30 y=10x - 5 47 ©Prep101 CALC 1000 Booklet Solutions Example 5. Subst. X=3 into f(x) and get (3) = √3 + 1 = 2 Point (3,2) = ′ (3) = lim (√3 + ℎ + 1 − 2)(√3 + ℎ + 1 + 2) ℎ(√3 + ℎ + 1 + 2) ℎ→0 = lim (3 + ℎ + 1 − 4) ℎ→0 ℎ(√3 ℎ = lim ℎ→0 ℎ(√3 + ℎ + 1 + 2) 1 = lim ℎ→0 (√3 = + ℎ + 1 + 2) + ℎ + 1 + 2) 1 4 1 1 3 − 2 = ( − 3) = − + 2 4 4 4 1 5 = + 4 4 1. ( ) = √ ′ ( ) = lim (+ℎ)−() ℎ→0 ℎ √+ℎ−√ ℎ ℎ→0 = lim (√+ℎ−√)(√+ℎ+√) ℎ(√+ℎ+√) ℎ→0 = lim = lim +ℎ− ℎ→0 ℎ(√+ℎ+√) = lim ℎ lim 1 ℎ→0 ℎ(√+ℎ+√) ℎ→0 (√+ℎ+√) = 1 2√ 48 ©Prep101 CALC 1000 Booklet Solutions N2. ( ) = ( + 1)2 = 2 + 2 + 1 ( + ℎ) = ( + ℎ + 1)2 = ( + ℎ + 1)( + ℎ + 1) = 2 + ℎ + + ℎ + ℎ2 + ℎ + + ℎ + 1 = 2 + 2ℎ + 2 + 2ℎ + ℎ2 + 1 ′ ( ) = lim (+ℎ)−() ℎ ℎ→0 = lim 2 +2ℎ+2+2ℎ+ℎ2 +1−( 2 +2+1) ℎ ℎ→0 = lim 2ℎ+2ℎ+ℎ2 ℎ ℎ→0 = lim ℎ(2+2+ℎ) ℎ ℎ→0 = lim(2 + 2 + ℎ) ℎ→0 = 2 + 2 3. ( ) = 2 − ( + ℎ) = ( + ℎ)2 − ( + ℎ) = 2 + 2ℎ + ℎ2 − − ℎ ′ ( ) = lim (+ℎ)−() ℎ→0 = lim 2 +2ℎ+ℎ2 −−ℎ−( 2 −) ℎ ℎ→0 = lim ℎ→0 = lim ℎ→0 ℎ 2ℎ+ℎ2 −ℎ ℎ ℎ(2+ℎ−1) ℎ 49 ©Prep101 CALC 1000 Booklet Solutions (2 + ℎ − 1) ℎ→0 1 = lim = 2 + 0 − 1 N4. ( ) = = 2 − 1 1 ( + ℎ ) = +2 ′ ( ) = lim ℎ→0 = lim +ℎ+2 (+ℎ)−() ℎ ℎ→0 = lim 1 1 1 − +ℎ+2 +2 ℎ 1(+2) 1(+ℎ+2) − (+2)(+ℎ+2) (+2)(+ℎ+2) ℎ ℎ→0 = lim +2−−ℎ−2 (+2)(+ℎ+2) ℎ ℎ→0 −ℎ 1 −1 lim (+2)(+ℎ+2) ( ) = (+2)(+0+2) = ℎ ℎ→0 1 5. ( ) = √ ′ ( ) = lim ℎ→0 = lim ℎ→0 = lim ℎ→0 (+2)2 1 √ + ℎ (+ℎ)−() ℎ ℎ→0 = lim ( + ℎ ) = −1 1 1 − √+ℎ √ ℎ √+ℎ √ − √√+ℎ √√+ℎ ℎ (√−√+ℎ)(√+√+ℎ) √√+ℎ (√+√+ℎ) ℎ 50 ©Prep101 = lim CALC 1000 Booklet Solutions −(+ℎ) √√+ℎ(√+√+ℎ ℎ ℎ→0 = lim −−ℎ √√+ℎ(√+√+ℎ) ℎ ℎ→0 −ℎ = lim 1 ( )= ℎ→0 √√+ℎ(√+√+ℎ) ℎ −1 = 3 = −1 2 2 −1 √√+0(√+√+0 = −1 √√(√+√) = −1 (2√) = −1 1 2 1 2 −3 2 2 2 N6. A) Find ′ () using the definition of the derivative if () = (2 − 3) . ( + ℎ) = [2( + ℎ) − 3]2 = (2 + 2ℎ − 3)(2 + 2ℎ − 3) ( + ℎ) = 4 2 + 4ℎ − 6 + 4ℎ + 4ℎ2 − 6ℎ − 6 − 6ℎ + 9 ( + ℎ) = 4 2 + 8ℎ − 12 − 12ℎ + 4ℎ2 + 9 And ( ) = (2 − 3)2 = (2 − 3)(2 − 3) = 42 − 12 + 9 ( + ℎ) − () ℎ→0 ℎ ′ () = lim 4 2 + 8ℎ − 12 − 12ℎ + 4ℎ2 + 9 − (4 2 − 12 + 9) ℎ→0 ℎ ′ () = lim ′ () 4 2 + 8ℎ − 12 − 12ℎ + 4ℎ2 + 9 − 4 2 + 12 − 9 = lim ℎ→0 ℎ ′ () 8ℎ − 12ℎ + 4ℎ2 = lim ℎ→0 ℎ ℎ(8 − 12 + 4ℎ) ℎ→0 ℎ ′ () = lim ′ () = lim (8 − 12 + 4ℎ) ℎ→0 51 ©Prep101 CALC 1000 Booklet Solutions ′ () = 8 ...
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