MATH140 Exam 2 - Sample Test 1 Detailed Solutions

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1 1. D. reate a first derivative number line MATH140 Eam - Sample Test 1 Detailed Solutions cos -1 0 cos -1 cos 1 cos 1/ p + æp ö p æp ö ç è 4 ø ç è ø.. reate a second derivative number line. Remember that a number is only an inflection point if it changes concavity at that point f ''( ) ( ) ( )( ) 0,, Since f ''( ) does not change signs at 0, it is not considered an inflection point.. B. For parts A & B, look at the first derivative number line. Parts & D are false because the domain of the original function is all real numbers. Part E is false because the second derivative does not change signs around / /5 0 1 LionTutors 019

2 4. E. To find the absolute ma and min, find the critical numbers that are within the interval. Take the critical numbers, and the endpoints of the interval and plug into the original function. ( -)( ) -( )( 1) ( - ) -4- ( - ) - 4 ( - ) The critical numbers are 04,, however we only consider 4 since it is the only number in the interval. f () 9 f ( 4) 8 f ( 7) 49/ 5 ß Minimum ß Maimum 5. D. Simplify by multiplying by the conjugate lim * lim (4 - ) 1 lim lim lim LionTutors 019

3 6. E. Use the definition of the MVT. f( 16) - f( 1) (Note that you do not need to consider -4 because it is not in the interval) 7. D. This is an optimization problem. We want to minimize the distance between the graph and the point. First solve for y, then use the distance formula. Then take the derivative of the distance and set it equal to 0. Remember that you can square the distance equation before taking the derivative, to make it easier. This will not affect the answer. y 56-8 y 56-8 d ( - ) + ( ) d ( - ) d ( - ) ( -) / 7 LionTutors 019

4 8.. To find horizontal asymptotes look at the powers. Since the highest power is the same on top and bottom then divide the coefficients (don t forget the signs). To find vertical asymptotes, SIMPLIFY, then set the denominator equal to ( - )( 1+ ) f( ) ( + )( + 1) 41 ( - ) f( ) B. Use long division to divide the numerator by the denominator ( ) B. To determine which graph this is look at intercepts, horizontal asymptotes, vertical asymptotes. The graph has a H.A. at y 0, so we know the power on bottom must be bigger. This eliminates answer choice D. The graph has V.A. at 1 and -1. This eliminates are choices & E because those functions only have a V.A. at 1 (remember to simplify the function and set the denominator 0 to find V.A.) The function has a y-intercept at 0. Therefore when you plug in a 0 for, you should get 0 for y. This eliminates answer choice A. Therefore, by process of elimination this graph has to be answer choice B. 4 LionTutors 019

5 11. D. Find the linearization, then plug in.1 y- 10 -( -) y y L ( ) L(. 1) - (. 1) + 16 L( 1. ) E. To find dy calculate the differential. To find Dy calculate the actual change. dy d dy ( 8)( d) dy 8 ( )(. 5) dy 8 D y f( 5. )- f( ) é æ5 ö ù D y ê4ç -1ú-é4( ) -1ù ê ú ë û ë è ø û D y 4-15 D y 9 1. A. This is an application of the Mean Value Theorem. f( 10) - f( 8) ³ 10-8 f ( 10) - ³ 7 f ( 10) - ³ 14 f ( 10) ³ 17 5 LionTutors 019

6 14.. This is an application of the nd Derivative Test. Plug each of the critical numbers into the nd derivative. If you get a positive value then it is a local minimum. If you get a negative value then it is a local maimum. f ''( 0) -, so there is a local maimum at 0 7 f ''( / ), so there is a local minimum at / 4 f ''(- ) 5, so there is a local minimum at B. This is an optimization problem y y P + 4y æ00 ö P + 4ç è ø P P' y 00 0 y 10 P 0 ( ) ( ) P First simplify, then take the antiderivative. sin + 1 sin 1 1 csc + + sin sin sin So the antiderivative is - cot + 6 LionTutors 019

7 17. B. Take the antiderivative and then use the point to solve for c. The antiderivative of - sin is f( ) + cos cos( 0) æp ö æp ö p f ç ç + cos + 4 è ø è ø æp ö p + f ç è ø cos This is an optimization problem. Find a formula for volume, then take the derivative and set it equal to 0. V l w h V ( 18 -)( 18 -)( ) V ( 18 -) V ' ( 18 - ) ( 1) + ( )( )( 18 - )(-) V ' ( 18 - ) -4 ( 18 - ) 0 ( 18 -)[ ] 0 ( 18 -)( 18-6) 9, However, cannot equal 9. Since there is only 18 inches to work with, you cannot take away 9 inches from each side. 19. TRUE. This is the definition of Rolle s Theorem. Also note that since f is differentiable, then it must also be continuous according to the definition of differentiability. 0. FALSE. An absolute minimum can occur at the end point of the interval you are looking at. 1. FALSE. The second derivative also needs to change signs at that point. Look at the function f() ) as a countereample. 7 LionTutors 019

8 . It helps to create the 1 st and nd derivative number lines before trying to graph. a 0 0. a) - Intercept(s): (0,0) b) y Intercept: (0,0) c) Vertical Asymptote(s): NONE d) Horizontal Asymptote(s): y 0 (since the power on bottom is bigger) e) Increasing: (-1,1) + f) Decreasing: (-,1)U(1, ) g) Relative maimum(s): h) Relative minimum(s): i) oncave Up: (-, 0) U (, ) j) oncave Down: (-,- ) U (0, ) - 0 k) Inflection points: -, 0, 8 LionTutors 019

9 MATH140 Eam - Sample Test Detailed Solutions 1.. Remember that finding the linearization just means finding the equation of the tangent line. Find the point by plugging 0 into the original function and find the slope by taking the derivative and plugging in 0. 1 Point: ( 0, ) 5 Slope: f( ) ( + ) -1 5 / 1 - ( + 5) -/ f '( 0) ( 5) Line: y ( - ) y y L ( ) A. Use differentials to approimate dy, the change going from 10 to 10.. Remember that in this case your d 0.. Add this to f ( 10) dy f '( 10) d dy -8(. ) dy -4. to get the answer. 9 LionTutors 019

10 f( 10. )» f( 10) + dy f ( 10. )» 0 + (-. 4) f ( 10. )» Take the derivative, using the quotient rule. Then set the top and bottom equal to 0. Remember that 0 cannot be a critical number since it is not in the domain of the original function. ( - ) f( ) / -1 / é ù -1 / / 1 ( -) -( -) * 1 ( - ) ê -( - ) ú ë û ( -) ( -) 1/ 1/ 1 1-0Þ Þ 6-0Þ 4. D. Find the critical numbers and set up a first derivative number line ( - 1)( + ) 0, 1, E. Find the critical numbers in the interval. Take those numbers and the endpoints of the interval and plug into the original function. sin cos + cos 0 cos (sin + 1) cos 0 sin -1 p, p p f ( 0) 0 æp ö f ç è ø æp ö f ç - 1 è ø f ( p ) 0 10 LionTutors 019

11 6. E. Use the second derivative test to answer this question. Whenever the nd derivative is positive at a critical number, there is a local minimum. When the nd derivative is negative at a critical number, there is a local maimum. If the nd derivative is equal to zero, then you cannot tell. f ''( 0) 0 f ''( ) 8( - ) -16 f ''( 6) 4( 6) 144 ß local maimum ß local minimum 7.. To find inflection points look at the nd derivative. Remember that concavity must change at that point in order to be an inflection point f ''( ) ( 1- ) ( - )( + ) 0, 1, f( ) has two inflection points (at 1 and -1) 8. A. Answer choice A is the false statement because -4 is not in the domain, so it cannot be considered a critical number. 9. B. Use the definition of the MVT to answer this question. 11 LionTutors 019

12 f() 1 - f( 0) 1-0 (-4)-(-5) E. To find VA set the denominator equal to zero. To find HA look at the highest power on top and bottom. V.A. ( ) H.A. The powers are the same on top and bottom so we take the coefficients and divide. The coefficient on top is - and the coefficient on bottom is 4 (because we have squared). -/4 -. So the H.A. is y -1/ 11.. To take the limit as goes negative infinity, look at the most dominant terms on top and bottom. lim lim A. To find V.A., set the denominator equal to D. Take the second derivative and set up a second derivative number line. 1 LionTutors 019

13 - sin + f ''( ) -cos - cos 0 p p, + p p 14. B. First simplify the function, then take the antiderivative. 1 t f() t - t t f() t t t So the antiderivative is - t Which simplifies to t - + t 15. D. Take the anti-derivative twice, solving for each time. 1 LionTutors 019

14 at () 6t-4 vt () t - 4t () - () vt () t - t+ 4 5 st () t - t + 5t st () t t t s() s() TRUE. First note that since the derivative eists, the function is continuous. If the derivative is non-zero, then it means the function is always increasing or always decreasing. Therefore it cannot have the same y-value twice. 17. TRUE. Every function has an absolute maimum and minimum on a closed interval. 18. FALSE. Functions do not necessarily have an absolute maimum on an open interval. 19. FALSE. If f and g are both increasing, it means that f 0 () > 0 and g 0 () > 0, however it does not necessarily mean that f and g are greater than 0. When you computer fg, you have to use the product rule, so if the original function is less than 0, then the product may also be less than 0, and thus not necessarily increasing. 0. This is an optimization problem. Overall we want to minimize cost. 14 LionTutors 019

15 y æ 8 ö + ç è ø ()( ) V 8 y 8 8 y ' y ( ) y 1. Take the limits as goes to infinity and negative infinity lim lim lim lim ( ) lim lim lim lim So the H.A. are y 1, and y LionTutors 019

16 a) Intercept(s): (0, 0) b) Vertical Asymptote(s): NONE c) Horizontal Asymptote(s): y 0 d) Graph crosses the horizontal asymptote at 0 (set the equation to the asymptote to find this) e) Increasing: (-1,1) _ + _ f) Decreasing: (-, -1) U (1, ) g) Relative maimum(s): (1, 1/ ) -1 1 h) Relative minimum(s): (-1, -1/) i) oncave Up: (-,0)U(, ) _ + _ + j) oncave Down:(,- )U(0, ) k) Inflection points: -, 0, LionTutors 019

17 . Looking at the graph of the derivative, create a first derivative number line by looking at where the graph is positive and negative. That is, where the graph is above or below the ais To create the second derivative number line, look at where the graph is increasing and decreasing increasing: (-,-1)U(-1,1)U(4, ) decreasing: (1,4) concave up: (-1,0)U(, ) concave down: (, -1)U (0, ) local maimum: 1 local minimum: 4 inflection point: -1, 0, This is an optimization question. Overall you are trying to minimize cost. 10*pr + 8*prh V pr h 0pr + 16prh 0p pr h 0 0 0pr + 16pr h r r -1 0pr + 0pr 0p ' 40p r- r 0p 0 40p r - r 0p 40p r r 0p 40pr 8 r r 0 h 5 () 17 LionTutors 019