BARUCH COLLEGE MATH 07 FALL 007 MANUAL FOR THE UNIFORM FINAL EXAMINATION The final eamination for Math 07 will consist of two parts. Part I: Part II: This part will consist of 5 questions. No calculator will be allowed on this part. This part will consist of 0 questions. The graphing calculator is allowed on this part. There may be a few new problem types on the eam that are not similar to the problem types that appear in this manual. If such problems appear, they will be similar to problems that you have seen during the semester. GRADING: Each question will be worth points. Anyone who gets 4 or 5 questions correct will be assigned a grade of 00. No points are subtracted for wrong answers. CONTENTS OF THIS MANUAL: One page showing the sample questions that correspond to each section of the current calculus tetbook. (When a section has been covered in class, the list indicates the problems that can be used in studying for the eam that includes that section during the semester.) For each section, Part I type questions are listed first; then Part II questions are listed. Problems numbered 6 and higher (plus F5) are Part II problems. TI-89 Facts for the Uniform Final Eamination. (This portion indicates the minimal calculator knowledge needed.) Five sample final eams, A to E. Additional sample problems. Answers to the sample eams and additional problems.
Tetbook Sections Corresponding to Sample Uniform Final Eam Questions Fall 007 Tetbook: Larson and Edwards, Brief Calculus, an Applied Approach, Seventh Edition, Houghton Mifflin, 006. Section Problems.5 B, B9.6 A, C, D7, E6, F4.6 A, A, B, B4, C4, C5, D8, E7, E8, F, F5; D8, F5. A0, A, B, B, C7, D5, D6, E5; C4, E. A9, B0, C6, D4, E4, F, F6; F6. A, B8, C5, F6; A0, D, E6, E7.4 A6, A7, B7, B8, C, C4, D, D, E, E; F9.5 A8, B9, C5, D, E, F7.6 F7.7 A6, B, B4, C0, D0, E5.8 A7, C, D, E7, F8. A, A, D4, E, E, F9; B6. A4, B, B4, C8, C, C, D6, D8, E; A6, C0, E. A5, B, B5, D9, E4, E6; A7, D0.4 A0, B7, B0, C4, D5, D4, E0.5 C7, D5, E, E; A, C.7 A.8 A, B9, B, C5, D7, E; C5 4. A5, B6, C, D0, E0, F, F0; A, B, C6, D9, D5 4.5 A4, B5, C, D9, E9; D4, F8 5. A8, B6, E8 5. & 5. A9, D 5.4 A9, C, C, D, E; B8, C8, E4 5.5 B5, D; A8, B7, C7, D, E5
TI-89 FACTS FOR THE UNIFORM FINAL EXAMINATION The following information represents the minimal calculator usage students should be familiar with when they take the uniform final eamination. Instructors are epected to provide much more information in class. Limits can be evaluated with the TI-89 by using the limit function whose synta is: limit(epression, variable, value). This function is found on the Home screen by using the F Calc key. The calculus menu that appears when F is pressed is shown in Figure. The limit function is selected by either pressing or using the down arrow cursor to highlight choice :limit and then pressing ENTER. Eample : Find Lim 4 6 6 Solution: Select the limit function as indicated above. Complete the command line as follows and press ENTER: limit((-4)/(6-/()),,6) The answer is as shown in Figure. Eample : Find Lim + 0 Solution: In order to use the calculator for this problem it is best to think of as a single letter such as t. That is, Lim + t t 0 t Now enter on the command line the following: limit((/(+t)-/)/t,t,0). The result is -/ as shown in Figure.. Note that eample could have been done without a calculator if the limit requested was recognized as the definition of the derivative of f() = / for which f N() = -/. Figure Figure Figure The synta for the solve command and for finding the first or second derivative is: solve(equation, variable) d(epression, variable) (first derivative, f N(), where f () = epression) d(epression, variable, ) (second derivative, f O(), where f () = epression) The solve command is obtained by pressing the F (Algebra menu) key in the Home Screen as shown in Figure 4 and then pressing ENTER to select choice :solve. The differentiation operator d is choice on the Home screen calculus menu shown in Figure. It can be more easily accessed by pressing the yellow nd key and then 8 (the d appears above the 8 in yellow). The purple D appearing above the comma key cannot be used for this purpose. Figure 4
If an answer such as e or 5/6 appears when a decimal answer is desired, the command can be repeated with the green diamond key pressed before pressing ENTER. Eample : The demand function for a product is p = 0 - ln(), where is the number of units of the product sold and p is the price in dollars. Find the value of for which the marginal revenue is 0. Solution: The revenue function is R() = p = (0 - ln()). The marginal revenue is the derivative, RN(). This can be found by hand or by using the calculator command d(*(0-ln()),) obtained by pressing the keys nd 8 ( * ( 0 - nd ) ), ) ENTER as shown in Figure 5. The result is RN() = 9 - ln(). To find the value of for which the marginal revenue is 0 the equation 9 - ln() = 0 must be solved. Figure 5 Press F (Algebra) ENTER as shown in Figure 4 and then complete the command line as: solve(9-ln()=0,) Pressing ENTER produces the result e 9. Since a decimal answer is desired, repeat the command by first pressing the green diamond key before pressing ENTER. The final answer is 80.08 as shown in Figure 6 and can be rounded off to 80. 0 Eample 4: The function ht () =. t has eactly one + e 0 0 + 5 inflection point. Find it. Solution: Recall that an inflection point occurs if ho(t) = 0 and the concavity changes sign. So first the second derivative is found with the command d(0/(0+e^(-0.t+5)),t,) where e^( is obtained by pressing (e appears in green over the key). Figure 7 is the result. Recall that to see the entire result the up cursor arrow button must be pressed and then the right cursor arrow button must be pressed until the rest of the result on the right can be seen as shown in Figure 8. Thus, t t. (. ) ((. ). ) h"( t) = 445 9 057 057 4 84 t 5 ( 0( 057. ) + e ) Move the cursor down to the command line again. The easiest way to solve ho(t) = 0 is as follows. First press F (Algebra) and select choice so that the command line now only shows solve( Press the up cursor arrow to highlight the epression for ho(t) and press ENTER. The command line now shows the end of solve(above epression Complete the command by entering =0,t)ENTER The answer appears in Figure 9. Since only one answer appears and it is known that an inflection point eists, there is no need to check to see if the concavity changes at t = 6.974. It only remains to find the value of the original function at t = 6.974. To do so, enter 0/(0+e^(-0.*6.974+5)) on the command line to obtain h(6.974) =.5 as shown in Figure 0. Therefore, the inflection point is (6.974,.5). Figure 6 Figure 7 Figure 8 Figure 9 Figure 0
Eample 5: 0.0559 and.788 are the only critical numbers for f() = e 5 ln(/). Determine if the critical point (0.0559, -4.7) is a relative minimum, a relative maimum or neither. Solution : Recall that if f O(0.0559) is positive the critical point is a relative minimum, if it is negative the point is a relative maimum, and if it is 0 the first derivative test must be used. The key with the symbol * on it means when or such that. Now enter the following command: d(e^(5)*ln(/),,)*=0.0559 Since f O(0.0559) = -04.9 is negative as shown in Figure, the critical Figure point is a relative maimum. Solution : A graph that clearly showed the point (0.0559, -4.7) would reveal what was true for the critical point. Press F for the y= screen. Enter the function y=e^(5)*ln(/) as shown in Figure. If F (Zoom) and choice 6:ZoomStd are selected (so that the and y values go from -0 to 0), the result is Figure. Notice that this reveals nothing about the critical point in question. So it is desirable to look at the point more closely. (If you are epert at using ZoomIn, this approach can be used instead of the one shown. Just make sure the new center chosen has a y value near -4.7.) Figure Figure A good start might be to select values of between - and and values of y between -6 and -4. So press F (Window) and enter the following values. min=- ma= scl= ymin=-6 yma=-4 yscl= as shown in Figure 4. Figure 4 Then press F (Graph). Figure 5 is the result. Now press F (Trace) and then press the right cursor arrow a few times and observe values of and y for each point on the graph shown. Clearly the high point is the critical point and therefore the critical point is a relative maimum. Figure 5 You should also be familiar with using F5 (Math) in the graph window to determine the eact location of a relative maimum or relative minimum displayed on the graph. For the graph shown in Figure 5 press F5 (Math) to obtain the menu shown in Figure 6. Select choice 4:Maimum. Figure 6
In response the Lower Bound? request, just use the cursor movement arrows to move the blinking cursor to the left of the maimum and press ENTER. Then, for the Upper Bound? request, move the blinking cursor to the right of the maimum and press ENTER. Figure 7 is the result and indicates that (0.0559, -4.709) is a relative maimum. Figure 7 Eample 6: Given f() = 4 + 4-4 - 6 + 5 i) Find the critical numbers. ii) Find the critical points iii) All of the following are graphs of f() in different graphical windows. Which graph most accurately portrays the function (shows its relative etrema, asymptotes, etc.)? a) b) c) d) e) Solution: i) The critical numbers are the solutions to 0 = fn() = 4 + 4-48 - 6. Enter the command solve(4^+4^-48-6=0,) into the calculator as shown in Figure 8. The critical numbers are = -.45, -.06 or.479 Figure 8 ii) For the critical points the value of y is needed for each of the critical numbers found. First enter the function into y: y=^4+4^-4^-6+5. Return to the Home screen by pressing the HOME key and press y(-.450) ENTER in order to find y = f (-.450) = -540.64. Repeat this for the other critical numbers. (Pressing the right cursor arrow removes the command line highlight and positions the cursor at the end of the line. Then the backspace key ³ can be used to eliminate the -.45. Then enter the net critical number followed by ).) The critical points are (-.45, -540.64), (-.06, 0.45) and (.479, -86.94). iii) A graphing screen should be used that includes the critical points found. The values shown should include all values between - and. The y values shown should include all values between -540 and. A reasonable window to choose would thus be min=-5 ma=5 scl= ymin=-5500 yma=500 yscl=500 The graph shown in Figure 9 is the result. So the answer is d. Figure 9
The integral sign appears in yellow over the key labeled 7. It can also be accessed in the home screen by pressing F Calc : *( integrate. The synta is: *(4^,) means 4d 4 *(4^,,,) means and returns 80. and returns 4 (It is assumed that you will add the constant: 4 + c.) d Eample 7: Find the area bounded between y = 4-8 and y = - 9. Solution: We know the parabola opens upward and has (0, -9) as its verte. (If it is not clear to you as to why that is the verte, notice 0 = yn = Y = 0.) Enter the first function into y: ^4-8^ so that it is available for graphing. Find where its derivative is 0: solve(0=d(y(),),) Y = -, 0, Find the corresponding values of y: y(-) is -6, y(0) is 0 and y() is -6. The critical points are (-, -6), (0, 0) and (, -6). Enter the parabola in y: ^-9 Choose a graphing window that shows the verte of y and the critical points of y: min = -4 ma = 4 scl = ymin = -0 yma = 0 yscl = 5 Figure 0 is the result. Net we find the points of intersection. Figure 0 solve(y()=y(),) Y = -, -,, We can use either y or y to find the values of y for these 4 values of (e.g. y(-) is 9). The 4 points of intersection are (-, 9), (-, -7), (, -7) and (, 9). There are no other points of intersection and these are the 4 points of intersection that appear in Figure 0. So we wish to find the area shaded in Figure. Notice that the parabola is on top between - and -, the 4 th degree function is on top between - and, and the parabola is on top between and. The area between - and - is (( 9) 4 ( 8 )) d 4 = ( + 0 9 ) d and this is *(-^4+0^-9,,-,-) = 04 5 The area between and is (( 8 ) ( 9)) d = ( 4 0 + 9 ) and this is *(^4-0^+9,,-,) = 76 5 d The area between and is = Figure (( 4 9) ( 8 )) d ( 4 + 0 9 ) d and this is *(-^4+0^-9,,,) = 04 5 Therefore the desired area is 04 76 04 784 4 5 + 5 + 5 = 5 = 5 5 = 5. 667
MATRICES Storing a matri in a TI-89 calculator results in the calculator keeping that matri in memory until it is erased. This is true whenever any result is stored in a variable. But from a practical point of view, this is usually only done for matrices. Hence, it is best if all variables are cleared after using the calculator with matrices. In the home screen, proceed as follows. Press F6 ( nd F) Then press ENTER. The screen shown below appears. The screen shown below appears. Press ENTER again. The home screen reappears. The TI-89 will now be used to enter the matri A =. Press APPS Press 6 (Data/Matri Ed.) Press (New) Notice that the word data is blinking. That indicates where the cursor is. Press right cursor Press (Matri) Press down cursor twice so that the cursor blinks in the variable rectangle. Press A (key with = on it)
Press down cursor,, down cursor, Press ENTER twice Press, ENTER Press 7, ENTER 5, ENTER, ENTER 4, ENTER Grey (-), 6, ENTER (The cursor movement keys can be used to go back to any entry that needs changing.) Press HOME Press A (purple alpha, =) Press ENTER Apart from going to the home screen, the last command (a, ENTER) was not needed. However, it does represent a useful way to display the final matri and confirm the fact that the matri was correctly entered into the calculator. Eample 8: Use the TI-89 to find the reduced row echelon form of A =, the matri that was just entered Solution: Matri A was entered into the calculator above. Go to the Home screen and clear the command line. Net press the catalog key. Then press, which has the letter R in purple above it (when using catalog it is not necessary to press the alpha key before pressing a letter). Scroll down with the down cursor until you reach rref. Position the solid pointer so that it points to rref as shown in the figure on the right. Press ENTER, alpha, a, ), ENTER to obtain rref(a). Your screen should look like the one shown on the right. The reduced row echelon form of A is Note: All commands can be entered by hand if that is desired. In this case pressing alpha, r, alpha, r, alpha, e, alpha, f, (, alpha, a, ) would be used.
5 Eample 9: Given the matrices A =, B = and C = find: a) A - b) B - c) A - B d) A - C Solution: Enter the three matrices A, B and C. It is very likely that you encountered the screen shown on the right when you tried to enter A. That is because you must first delete the previous matri entered that was also called A as was shown at the beginning of the section on matrices: F6 (i.e. nd F), ENTER, ENTER. To find A -, simply enter A^(-) on the command line of the home screen, i.e. alpha, 8, =, ^, (, (-), ),. The result appears on the right. Hence, A - = / 7 / 7 / 7 0 / 7 / 7 4/ 7 To find B - enter B^(-) in the same way. The figure shown on the right stating that B is a singular matri is the result. This means the matri B does not have an inverse. To find A - B simply enter that epression on the command line, i.e., alpha, 8, =, -,, alpha, 8, (, ENTER Hence, A - B = 7 7 5 8 5 7 44 5 To find A - C, enter that epression on the command line, i.e. alpha, 8, =, (, (-), ), *, alpha, 8, ), ENTER Hence, A - C = 8 4
SAMPLE EXAM A PART I: NO CALCULATOR ALLOWED + for f( )= + for > A. Describe the intervals on which is continuous. A) All real B) All real C) All real D) All real E) All in the interval (0,4) A. Find Lim 4 5 + 4 0 + + A) /4 B) 4 C) 0 D) E) 4 f ()= t t t + 4t + A. The function has vertical asymptotes A) t = only B) t = - only C) t = only D) t = and t = only E) t = -, t = and t = A4. Find the derivative of y = ln(5-9). ln( 5 9) 5 9 5 5 9 A) B) C) ln(5) D) E) 5 ln( 5 9) f e ( )= A5. If, then f N() = e A) B) e - C) e - ( - ) D) ( - ) E) e - ( - ) e A6. Find the derivative of y = + 5 + 7 0 + 9 ( 5 + 7) ( 5 + 7) A) B) C) D) E) 5 ( 5 + 7) 5 ( + ) ( 5 + 7) A7. Find the value of the derivative of y = 5( - 4) 4 at =. A) 040 B) 560 C) 0 D) 60 E) 960 f( )= A8. The derivative of is 9 9 A) B) -9 ( - ) C) D) E) -9
A9. Find the equation of the line tangent to the graph of y = - + at the point (, ). A) y = - B) y = 4-6 + 5 C) y = + D) + y = 5 E) y = + f( + ) f( ) A0. Evaluate for f () = + 5. A) 7 + B) + C) 5 + D) + 5 + E) 0 + 5 Lim + ( ) 0 A. Find. A) 0 B) 5 + C) 5 D) 4 E) undefined A. Find the critical numbers for g() = 4 -. A) 0, ± B) 0, ± C) ± D) only 0, - E) only 0 A. Given y =, find the value of the differential dy if changes from 4 to 6. A) 48 B) 4 C) 60 D) E) 0 A4. (0, 5) (i.e. = 0 and y = 5) is the only critical point of f () = 6 + 4 + 5. You do not have to verify this. Determine what is true of (0, 5). A) (0, 5) is a relative maimum. B) (0, 5) is a relative minimum C) (0, 5) is a saddle point D) (0, 5) is not a relative etremum E) no conclusion is possible A5. On the interval 0 < <, for the graph shown, A) f N() > 0 and f O() > 0 B) f N() > 0 and f O() < 0 C) f N() < 0 and f O() > 0 D) f N() < 0 and f O() < 0 E) None of the above A6. Find dy/d by using implicit differentiation if y + y = 5. 9 0 + 0 y + A) B) C) D) 0 - E) 0 - y A7. At what rate is the area of a circle increasing if the radius is increasing at the rate of 0 feet per minute when the radius is 6 feet long? (A = π r ) A) π ft /min B) 60 ft /min C) 0π ft /min D) π ft /min E) 0 ft /min
A8. If the marginal profit function is given by PN() = 00-4 and the profit is $00 when 0 units are produced, find the profit when 0 units are produced. A) $5 B) $00 C) $400 D) $50 E) $600 f( )= A9. Find the average value of on the interval [, 4] (i.e. # # 4). A) 7/ B) 7/ C) / D) 4/9 E) 4/ A0. Select the correct mathematical formulation of the following problem. A rectangular area must be enclosed as shown. The sides labeled cost $0 per foot. The sides labeled y cost $0 per foot. If at most $00 can be spent, what should and y be to produce the largest area? A) Maimize y if 40 + 0y = 00 B) Maimize y if 0 + 0y = 00 C) Maimize y if + y = 00 D) Maimize + y if y = 00 E) Maimize + y if 00y = 00 A. What is the average rate of change of f () = - 4 + on the interval [-, ]? A) -4 B) 4 C) -6 D) - E) f( )= + 4 A. Find the open intervals on which is decreasing. A) (-4, -) and (, 4) (i.e. < - and > ) B) (0, 4) (i.e. > 0) C) (-, ) (i.e. - < < D) (-4, 0) (i.e. < 0) E) (-4, 4) (i.e. all real ) A. The system of linear equations on the right + y - z = - has as its solution y - z = 0 - + y = A) only (,, ) B) only (-, 0, 0) C) only (,, ) D) no solution E) (- + z, z, z) 4 5 0 7 4 5 4 A4. If A =, B = and C =, find AB - 5C. 7 5 0 45 9 A) B) C) does not eist because AB does not eist. D) does not eist because AB and 5C have different dimensions. E) does not eist because all three matrices must have the same dimensions for AB - 5C to eist.
A5. When writing the system of equations shown in the form AX = B, - y = 8 what should A be? -5 + 7y + z = 4 y + z = 7 0 5 7 0 0 8 5 7 4 0 7 8 5 7 7 0 8 5 7 4 0 7 A) B) C) D) E) 8 4 7 PART II: CALCULATOR ALLOWED A6. The absolute maimum value of the function f () = - 6 + 5 on the closed interval [0, ], (i.e. 0 # # ) is A) -7 B) -7 C) 4 D) - E) 5 40 ()= + + 5 4 A7. Find the inflection point for gt t A) (4, 8) B) (.678, 4) C) (0,.9048) D) (.5546,.890) E) (0.78, 7.9808) A8. Find the area of the bounded region between f() = 4-6 and g() = 0 (the -ais). A) 6 B) C) 8 D) 4 E) 0 6 d A9. Evaluate. 4 4 8 6 A) 0 B) = 6.597 C) 8π = 5.7 D) E) Non-real result C ( )= + 9 A0. The cost in dollars of producing units of a product is given by for $ 0. Determine the value of when the marginal cost is 0. A) 0.8 B).4 C).5 D) 4. E) 5. A. Find the number of units to be produced in order to maimize revenue if the demand function is given 500 p = 500 + 0. 00 by. A) 46 B) 54 C).547E-0 D) 844 E) A maimum does not eist (false)
y = f( ) = 4 + A. What is the relative maimum of the function? A) (-, 4) B) (, 0) C) (0, 0) D) (, ) E) (,.6) A. Find the number of units that will minimize the average cost function if the total cost function is A) 6 B) 0 C) 0 D) 40 E) 80 C ( )= + 400 4 A4. Solve the system of equations: - 4y - z = - y + z = - y + z = A) (0, 0, ) B) ( + 7z, y + 5z, z) C) ( + 7z, y + 5z, ) D) ( - 7z, y - 5z, ) E) no solution 5 A5. If A =, find A - / 5 / / / 5/ / / 5 5 A) B) C) D) E) Does not eist
SAMPLE EXAM B PART I: NO CALCULATOR ALLOWED 4 B. If A =, find A. 4 9 4 8 7 4 6 9 4 8 6 A) B) C) D) E) 6 4 9 B. What is the slope of the line tangent to y = 8 at the point (, )? A) -/ B) C) D) 8 E) -8 B. If R = - + - is the revenue function, where is the amount spent on advertising, then the point of diminishing returns (point of inflection) occurs when A) = B) = - C) = 6 D) = - E) Not enough information is given to decide B4. If y =, find dy/dt when = 6 and d/dt = 4. A) / B) -/ C) 0 D) ½ E) undefined B5. Find the area of the bounded region between f() = 9 - and g() = 0 (the -ais). A) 0 B) 8 C) 6 D) 5/ E) 9/ B6. The marginal cost, in dollars, is given by CN() = + 00. If the fied overhead cost is $000, then the total cost of producing 0 items is A) $4500 B) $000 C) $00 D) $500 E) $400 B7. Select the correct mathematical formulation of the following problem. A farmer wishes to construct 4 adjacent fields alongside a river as shown. Each field is feet wide and y feet long. No fence is required along the river, so each field is fenced along sides. The total area enclosed by all 4 fields combined is to be 800 square feet. What is the least amount of fence required? A) Minimize y if 4 + 5y = 800 B) Minimize y if 4 + 5y = 00 C) Minimize 4 + 5y if y = 800 D) Minimize 4 + 5y if y = 00 E) Minimize 4( + y) if y = 800
B8. Find the average rate of change of the function f () = - + 7 + on the closed interval [, 4]. A) -4 B) 7 C) -6 D) -4 E) 4 B9. Let the profit function for a particular item be P() = -0 + 60-00. Use the marginal profit function to approimate the increase in profit when production is increased from 5 to 6. A) 0 B) 40 C) 55 D) 60 E) 70 B0. The sum of a number and twice another number is 0. Find the largest possible product. (Notice that your answer is not one or both of the numbers, it is the product.) A).5 B) 8 C) D).6 E) B. The graph of y = f () appears on the right. Estimate the points at which the absolute minimum and the absolute maimum occur on the interval [0, ], that is, 0 # #. A) absolute minimum: (0, 0); absolute maimum: (, ) B) absolute minimum: (0, 0); absolute maimum: (, 4) C) absolute minimum: (-, -4); absolute maimum: (, 4) D) absolute minimum: (4, -5); absolute maimum: (-4, 5) E) absolute minimum: (, ); absolute maimum: (, 4) B. Given f( )= + for < for find the values of the limits shown. Choice Lim f ( ) Lim f ( ) Lim f( ) + (A) - 4 undefined (B) (C) 0 undefined (D) 7 4 undefined (E) 7 and 4 7 and 4 7 and 4 B. Find the horizontal asymptote of f ( ) = 5 + 7 4 + 8 A) y = 5 B) y = 7 C) y = 0 D) y = /8 E) y = 5/7 f( )= B4. has as vertical asymptotes A) = - and = B) = only C) = 0 only D) = - only E) = and = - B5. Find the derivative of y = ln ( + ) 7 4( + ) ln( + ) A) 4 ln ( + ) 6 B) 4 ln ( + ) 7 C) D) E) 6 7 4 + 7 +
f( )= e + 5 7 B6. Find the derivative of. 5e 5 + 5 5 7 e e 5 5e + e 5 + 7 5 + 7 A) B) C) D) E) e + 5 e 5+ 7 B7. Find the derivative of y = ( ) A) B) C) D) E) ( 6 + ) ( ) ( ) ( 6 ) ( ) 6 ( ) ( ) B8. If g() = ( + )( - + ), then the slope of the line tangent to the graph of g() at the point where = - is A) 0 B) C) - D) - E) 8 y = + 0 B9. Find the slope of the line tangent to at (, 4). A) B) /8 C) 6 D) E) /8 f( )= B0. Find the derivative of. A) B) C) D) E) 6 6 6 f( + ) f( ) B. Evaluate for f () = + 4. + 8 A) B) C) 4 + 4 + D) 4 + E) + ( ) + 4 Lim + 5( ) + 4 0 B. Find. A) 4 B) + 9 C) D) + 4 E) 0 B. If y = +, find the value of the differential dy corresponding to a change in of d = 0. when =. A) 0. B) 0.5 C) 0.5 D). E) 6.5
B4. (0, 7) (i.e. = 0 and y = 7) is the only critical point of f () = 7-4 - 5 6. You do not have to verify this. Determine what is true of (0, 7). A) (0, 7) is a relative maimum. B) (0, 7) is a relative minimum C) (0, 7) is a saddle point D) (0, 7) is not a relative etremum E) no conclusion is possible B5. On the interval 0 < <, for the graph shown, A) f () > 0, f N() < 0 and f O() > 0 B) f () < 0, f N() < 0 and f O() > 0 C) f () < 0, f N() > 0 and f O() < 0 D) f () < 0, f N() < 0 and f O() < 0 E) f () > 0, f N() > 0 and f O() < 0 PART II: CALCULATOR ALLOWED 4 y = + B6. One critical number of the function is A).6 B) -0.48 C) 0.75 D) E) -.7 B7. Given the demand equation p = - + 5 and the supply equation p = +, find the consumer surplus. A) 0 B) 6 C) -8 D) 4 E) 44 f( )= 0 4 + B8. Find the average value of on the interval [, ] (i.e. # # ). A).8878 B) 6 C).949 D) - E) -.5 B9. Evaluate Lim 0 ( e ) A) B) 0 C) 4 D) undefined E) e f( )= + + 6 B0. Find the interval(s) on which is increasing. A) -8 < < B) < -8 or > C) - < < 8 D) < - or > 8 E) all real B. The maimum value of f () = e - is A) e -. 0.68 B) e..78 C) 0.5 D) -e. -.78 E) 0
B. Solve the system of equations: - y + z = - - y = 4 + z = - A) (-9/7, -9/7, 8/7) B) (,, ) C) (-9k + 7, -9k + 7, 8k + 7) D) (-9, -9, 8) E) No solution B. Solve the system of equations: - y + z - 7u = - y + z + 9u = - 6y + 7z + u = 8 A) (45/8, 0, 4/7, -/8) only B) (z + 45/8, 0, 4/7, -/8) C) (45/8, -, 4/7, -/8) only D) (y + 45/8, y, 4/7, -/8) E) no solution 0 0 0 0 0 0 0 0 0 0 0 0 4 B4. If A =, find the inverse of A, A -. 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 A) B) C) 0 0 0 0 / 0 0 0 0 / 0 0 0 0 / 4 D) E) Does not eist. 0 0 0 0 / 0 0 0 0 / 0 0 0 0 / 4 B5. Solve y 5 4 z = 7 4 A) (,, ) B) (, 9, -5) C) (6, 5, -) D) (8, 5, -9) E) No solution
SAMPLE EXAM C PART I: NO CALCULATOR ALLOWED f( ) = ln 7 C. Find the derivative of. 7 7 = + + ( 7) A) B) C) 7 7 + = + ( 7) 7 7 + = ( 7) 7 7 = ( + 7) D) E) 7 7 + = ( 7) C. Find the equation of the line tangent to the graph of y = e - 4 at the point where =. A) y = e + B) y = - C) y = - 4 D) y = 4-6 E) y = 5-8 y = f( ) = 4 C. For what value(s) of is the derivative of equal to zero or undefined? A) = 0 only B) = only C) = - onlyd) = -, only E) = -, 0, C4. The derivative of f () = (4-7) 4 is A) (4-7) 4 B) 4(4-7) C) 6(4-7) D) (0-7)(4-7) E) 7(4-7) f ( ) = 4 C5. The derivative of is A) B) C) D) E) 4 4 ( 4 ) 4 4 4 ( ) C6. What is the EQUATION of the line tangent to y = 4 at (, )? A) y = - 6 B) y = C) y = 48 D) y = + 8 E) y = 48-64 f( + ) f( ) C7. Given f () = 7, find. A) 4 + 7 B) 4 + ( ) C) 7 D) E) 4 + 7( )
C8. (0, ) (i.e. = 0 and y = ) is the only critical point of f () = 5 + +. You do not have to verify this. Determine what is true of (0, ). A) (0, ) is a relative maimum. B) (0, ) is a relative minimum C) (0, ) is a saddle point D) (0, ) is not a relative etremum E) no conclusion is possible C9. Given f () > 0, f N() < 0 and f O() > 0 on the interval 0 < <, pick the correct graph below. A) B) C) D) E) C0. Use implicit differentiation to find dy/d when = and y = - for + y + y =. A) -/ B) - C) 0 D) /6 E) undefined C. The area of a square is increasing at the rate of 7 square inches per minute. At what rate is the length of one of the sides increasing at the moment when the side has length 4? A) 7/4 B) 8 C) ½ D) 7/6 E) 7/8 C. Evaluate ( 4 ) d 0 4 A) 8 B) C) 4 D) 6 E) 0 C. Find the area of the region shown where y = 6. A) 6 B) 56 C) 5 D) 48 E) 78 C4. A farmer wishes to construct adjacent enclosures alongside a river as shown. Each enclosure is feet wide and y feet long. No fence is required along the river, so each enclosure is fenced along sides. The total enclosure area of all enclosures combined is to be 900 square feet. What is the least amount of fence required? A) 0 feet B) 0/ feet C)0 feet D) 70/ feet E) 60 feet C5. The height in feet above the ground of an object traveling straight upwards is given by s = t + 4t, where t is the time in seconds. What is the average velocity of the object between 0 and seconds? A) 58 B) C) D) 8 E)
C6. If f () = 00 and f N() = -6, estimate the value of f (4) for the function y = f (). A) 94 B) 88 C) 06 D) 4 E) 8 C7. Find the number of units to be produced in order to maimize revenue if the demand function is given by p = 600-0.0 for 5,000 # # 0,000. A) 5,000 B) 0,000 C) 5,000 D) 8,000 E) 0,000 C8. The reduced row echelon form that results from - y + z = - Gauss-Jordan row reduction of the system of - y - z = -9 linear equations shown appears below. - 4y + z = 0 5 0 5 0 0 0 0 Find all solutions to the system of equations. A) (, 5, 0) B) (5z +, z + 5, z) C) (,, 4) D) ( - 5z, 5 - z, z) E) (-4, 0, 5) C9. For the system of equations shown, what is the reduced row - y + z + 7w = echelon form that results from Gauss-Jordan row reduction? - y + z + 9w = - 6y + 7z + w = 8 7 0 0 0 0 0 7 0 0 0 0 0 A) B) C) 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 D) E) 0 0 0 0 0 0 0 0 0 5 0 4 4 0 C0. Perform the following matri operation:. 8 4 8 7 8 6 4 4 4 8 5 0 8 8 A) B) C) D) E) 8 9 4 8 5
C and C. For the function f () = 4 - - C. Find all critical points (, y). A) (, ) B) (, -) C) (, 0) D) (-, -4) E) (0, -) C. Find the absolute etrema on the closed interval [0, 5] (i.e 0 # # 5). A) minimum is - and maimum is B) minimum is -7 and maimum is - C) minimum is -7 and maimum is D) minimum is -7 and maimum is 5 E) minimum is -0 and maimum is 0 f( )= C. In order to make the function continuous at = we must define A) f () = 0 B) f () = C) f () = D) f () = E) f () = C4. Find Lim + 5 7. + 4 + A) /8 B) / C) 4 D) -7/ E) 0 C5. = is a vertical asymptote of y = f(), whose graph appears on the right. Find Lim f ( ) A) 0.4 B) C) 0 D) 4 E) - 4 PART II: CALCULATOR ALLOWED f( ) =. 8e 0 C6. The function has eactly two points of inflection. The value of for one of these points of inflection is A) 8 B) -6 C) -0.8 D) 0 E) 5. 6 C7. Find the area of the region bounded by y = and y = - +. A) 60/ B) C) 55.456 D) 0.85 E) - C8. Find the average value of f() = e on the interval [0, ] (i.e. 0 # # ). A) 5.598 B) 6.799 C).995 D) 6.799 E) 0.5 C9. For what open intervals is f () = + 0 + 5 increasing? A) < -0 or > 0 B) > 0 only C) -0 < < 0 D) < -0 only E) < 0 only
C0. Find the value of the relative maimum for the function f () = - 5. A) 5 = -.87 B) 5 =.87 C) 0 5 = -.6 D) 0 5 =.6 E) C ( ) = 0. 0 + + 400 C. Given the total cost function, find the value of for which the average cost is a minimum. A) -5. B) 5. C) 78.594 D) -78.594 E) 500 6 8 C and C. Given C = and D = 6 5 7 C. Find CD -. 5 7 A) B) C) D) E) 0 7 4 C. Solve CX = D. 5 7 9 4 4 6 9 8 A) B) C) D) E) 0 7 4 9 4 4 ( C4. Find Lim + ) + + 0 6 9 8 4 4 4 4 6 9 6 6 9 6 9 9 6 6 A) 0 B) Does not eist (undefined) C) + D) + E) + C5. The side of a square is measured and found to be 6.78 inches long with a possible error of 0.0 inches. Use a differential to estimate the possible error that might result when computing the area of the square. A) 0.0009 B).7905 C) 0.4068 D) 0.0 E) 0.04
SAMPLE EXAM D PART I: NO CALCULATOR ALLOWED D. Solve the system of equations. + z = + y - z = 0 y - z = 8 A) (, 7, ) B) ( - z, 7 + z, 0) C) ( - z, 7 + z, z) D) ( - z, 7 + z, ) E) No solution eists D. Perform the following matri operation: [ 4 ] 5 0 4 6 5 0 4 5 0 A) [ 4 ] B) 6 6 6 C) 4 D) 6 6 6 E) Undefined 4 6 4 Lim + D. Evaluate. A) 0 B) - C) / D) /4 E) Undefined f ( ) = + 7 D4. Find the open intervals on which is increasing. A) (-4, -) or (, 4) (i.e. < - or > ) B) (, 4) only (i.e > only) C) (-, ) (i.e. - < < ) D) (- 4, -) only (i.e. < - only) E) (-4, 4) (i.e. all real ) D5. Of all numbers (positive, negative and zero) whose difference is 4, find the two that have the minimum product. One of the numbers is: A) 4 B) - C) - D) E) - D6. The graph of y = f () appears on the right. Estimate the points at which the absolute minimum and the absolute maimum occur on the interval [-, 0], that is, - # # 0. A) absolute minimum: (-, -4); absolute maimum: (, 4) B) absolute minimum: (0, 0); absolute maimum: (, 4) C) absolute minimum: (-, -4); absolute maimum: (0, 0) D) absolute minimum: (4, -5); absolute maimum: (-4, 5) E) absolute minimum: (-, -); absolute maimum: (0, 0)
D7. The function defined by f( )= 9 if 4 5 if 4 < < 5 0 if 5 A) is continuous for real B) is discontinuous at = 4 only C) is discontinuous at = 5 only D) is discontinuous at = 4 and = 5 E) is discontinuous for 4 < < 5 7 D8. Find the horizontal asymptote of f( )= 5 A) y = 0 B) y = 7/5 C) y = -7/ D) y = 7 E) there is no horizontal asymptote D9. Find the derivative of y = 5 ln( - 7). 5 ln( 7) 5 7 0 7 A) B) C) 5 ln() D) E) 0 ln( 7) D0. Find the derivative of f () = 4e -. A) -e B) e - C) -e - D) 4e - E) 4e - f( )= 6 + D. The derivative of is 6 ( + ) A) B) C) D) E) ( - 6)( + ) + + 6 ( + ) + + D. Find the slope of the line tangent to the graph of the function f () = ( + 5) at =. A) 4 B) 6 C) 4 D) 0 E) 0 y = ( 6 + 7) D. Find the slope of the line tangent to at =. 5 9 6 6 A) B) C) 45 D) E) 9 5 g ( )= + D4. If, then gn() is A) + B) + C) + D) ( + ) E) +
f( + ) f( ) D5. Evaluate for f () = 4 + 7. +4 A) B) C) 7 + 4 D) 8 + 7 + 4 E) + 4( ) + 7 D6. Find Lim + ( ) + 7. 0 A) + 7 B) + 0 C) D) 4 E) 0 D7. Given y = + 4, find the value of the differential dy if changes from to 5. A) 6 B) 6 C) 4 D) E) - D8. (0, 6) (i.e. = 0 and y = 6) is the only critical point of f () = 4 6 + 4 + 6. You do not have to verify this. Determine what is true of (0, 6). A) (0, 6) is a relative maimum. B) (0, 6) is a relative minimum C) (0, 6) is a saddle point D) (0, 6) is not a relative etremum E) no conclusion is possible D9. On the interval 0 < <, for the graph shown, A) f N() > 0 and f O() > 0 B) f N() > 0 and f O() < 0 C) f N() < 0 and f O() > 0 D) f N() < 0 and f O() < 0 E) None of the above D0. The slope of the line tangent to the curve + y = 4 at the point (-4, ) is A) 4/ B) -4/ C) /4 D) -/4 E) D. If y = - 6 + 4, find d/dt when = and dy/dt = 60. A) ½ B) C) 4 D) 0 E) 800 D. Find the average value of f() = - 5 on the interval [-, ] (i.e. - # # ). A) -/ B) / C) D) - E)
D. In the graph on the right the parabola is f() = - 9 and the function whose graph turns times is g() = 4-8. The total area of the three shaded regions combined is A) ( f( ) g( )) d+ ( g( ) f( )) d+ ( f( ) g( )) d B) ( g ( ) f( )) d+ ( f( ) g ( )) d+ ( g ( ) f( )) d C) ( f( ) g( )) d D) E) None of the above ( g ( ) f( )) d D4. Select the correct mathematical formulation of the following problem. A rectangular area must be enclosed as shown. The sides labeled cost $ per foot. The sides labeled y cost $8 per foot. If the area must equal 900 square feet, what should and y be in order for the cost to be as small as possible? A) Minimize y if 4 + 6y = 900 B) Minimize y if + 8y = 900 C) Minimize 4 + 6y if y = 900 D) Minimize + 8y if y = 900 E) Minimize + y if y = 900 D5. For a production level of units of a commodity, the cost function in dollars is C = 00 + 400. The demand equation is p = 00-0.05. What price p will maimize the profit? A) $00 B) $50 C) $900 D) $500 E) $6000 PART II: CALCULATOR ALLOWED 4 5 D6. If A =, find the inverse, A -. 5/ 9 / 9 4/ 9 / 9 5/ 9 4/ 9 / 9 / 9 5/ 9 / 9 4/ 9 / 9 A) B) C) D) E) Inverse does not eist / 9 / 9 4/ 9 5/ 9 y = 7 z 4 7 5 4 D7. Solve. 7 4 A) B) C) D) E) 4 0
+ + D8. Find Lim A) 4 B) -4 C) - D) E) 0 D9. Find the critical numbers for f( ) = 6 e 0, 000 A) 0.48 only B) 0.48 and.94 C) D) 0.84 only E) 0.84 and 5.494 00, 000 D0. If dollars are spent on advertising, then the revenue in dollars is given by R = 5 0. 005, + where 0 # # 5000. How much is spent on advertising at the point of diminishing returns (point of inflection)? A) $0,576 B) $00,000 C) $50,086.60 D) $000 E) There is no point of diminishing returns D. Given the demand equation p = - + 0 and the supply equation p = + 5, find the producer surplus. A) 50 B) 5 C) -7.5 D).5 E) 6.5 D. Evaluate 8 4 7 d A) Non-real Result B) -4 C) -0.8 D) -0.6064 E) 6 D. The total profit of a company since it started business is given by P(t ) = 500t t 50, 000, where t is the number of months since the company started business. Find the average rate of change in profit with respect to time between months after the company started and 7 months after the company started. A) 5997.6 B) -4,506.6 C) 998.57 D) 499.8 E) 499. D4. The demand function for a product is p = 00-0. ln() where is the number of units of the product sold and p is the price in dollars. Find the value of for which the marginal revenue is one dollar per unit. A) 97.45 B) 0.00007 C) 88.9 D) 90.5 only E) 0.0 and 90.5 D5. = -/. -0. is a critical number for f( )= e. The critical point (-0., -0.497) is A) a relative minimum B) a relative maimum C) neither a relative maimum nor minimum D) a point of inflection E) a saddle point
SAMPLE EXAM E PART I: NO CALCULATOR ALLOWED E. The critical numbers for y = 4 - are A) = 0 only B) = -, 0, C) = -, only D) = - only E) = -4, 4 only E. If y = + 5 +, find the value of the differential dy corresponding to a change in of d = 0.4 when = 5. A) 76.4 B) 5.4 C) 5 D) 0 E) 6 E. (0, ) (i.e. = 0 and y = ) is the only critical point of f () = - 5 4-6. You do not have to verify this. Determine what is true of (0, ). A) (0, ) is a relative maimum. B) (0, ) is a relative minimum C) (0, ) is a saddle point D) (0, ) is not a relative etremum E) no conclusion is possible E4. On the interval 0 < <, for the graph shown, A) f () > 0, f N() < 0 and f O() > 0 B) f () < 0, f N() < 0 and f O() > 0 C) f () < 0, f N() > 0 and f O() > 0 D) f () < 0, f N() < 0 and f O() < 0 E) f () > 0, f N() > 0 and f O() < 0 E5. If 5 - y + y =, find dy/d by using implicit differentiation and evaluate dy/d at the point (0, ). A) - B) -5/4 C) 0 D) 6 E) Does not eist. E6. Find the point(s) of inflection (, y) for the function f () = - 9 + 5 A) (, 7) only B) (5, -5) only C) (, 7) and (5, -5) D) (, -9) E) (, -) E7. The side of a square is increasing at the rate of feet per minute. Find the rate at which the area is increasing when the side is 7 feet long. A) 8 ft /min B) 4 ft /min C) 49 ft /min D) 8π ft /min E) 49π ft /min E8. If the marginal probability is given by PN() = 00 - and the profit that results from producing 0 items is $600, find the profit that results from selling 0 items. A) $800 B) $900 C) $50 D) $400 E) $900 E9. Find the area of the region shown where y = 4. A) 40 B) 8 C) 8 D) 04 E) 80
E0. Select the correct mathematical formulation of the following problem. A farmer wishes to construct adjacent fields alongside a river as shown. Each field is feet wide and y feet long. No fence is required along the river, so each field is fenced along sides. The farmer has 900 feet of fence available. What should the dimensions of each field be in order to maimize the area of each field? A) Maimize + 4y if y = 900 B) Maimize + 4y if y = 00 C) Maimize y if ( + y) = 900 D) Maimize y if + 4y = 900 E) Maimize y if + 4y = 00 E. Find the maimum profit for the profit function P() = - + 0 -. 5+ 9 A) 0 B) 9/ C) D) 7/4 E) 67/8 E. If the derivative of g() is gn() = ( - ) ( - ) then the function is A) decreasing on (-4, 4) B) decreasing on (-4, ) and increasing on (, 4) C) decreasing on (-4, ) and increasing on (, 4) D) increasing on (-4, ) and decreasing on (, 4) E) decreasing on (-4, ), increasing on (, ) and decreasing on (, 4) E. C = 0.5 + 5 + 5000 represents the total cost of producing units. The production level that minimizes the average cost is A) = 5000 B) = 5 C) = -00 D) = 0,000 E) = 00 E4. The augmented matri shown represents the solution of a system of equations in the variables (, y, z, w). What is the solution? 5 A) (-, 0, 4, 9) B) + 0 y,, z, 7 8 z + 0 z C),, 9, z D) (- + 7z, 0-8z, z, 9) E) No solution 7 8 9 0 7 0 0 8 0 0 0 0 0 9 0 0 0 0 0 E5. Find the reduced row echelon form that results 4 + y = 6 from using Gauss-Jordan row reduction. 6 + y = 9 0 0 4 / / 0 / A) B) C) D) E) 0 0 5 0 0 0 0 0 0 0 0 0 0 4 E6. Determine the values of for which f( )= is continuous. 5+ 4 A) All real B) All real 4 C) All real D) All real, 4 E) All in the interval (0,5)
E7. Find the horizontal asymptote for the graph of y = + A) y = 0 B) y = C) y = D) y = E) y = - + E8. The graph of the function f( )= has as its vertical asymptote(s) + A) = - only B) = only C) = and = - D) No vertical asymptotes E) y = only E9. If y = ln (with > 0), then dy/d equals A) - ln B) C) ln D) + ln E) - ln f( )= e + 5 E0. If, then f N() = + e 5 + e 5 A) B) (6 + 5) C) e 6 + 5 D) e 6 + 5 (6 + 5) E) e 6 + 5 ( + 5) E. Find f N(-) for f ( ) = + A) 5 B) -5 C) D) - E) - E. Find the value of the derivative of y = (4-0) at =. A) 88 B) 88 C) 04 D) 96 E) 4 E. Find the derivative of f( )= +. A) B) C) D) E) + + + E4. The equation of the line tangent to the curve y = - + 7 at the point (, 8) is A) y = 4 B) y = 6 C) y = 4 + 4 D) y = 0 E) y = -4 + 8 f( + ) f( ) E5. Evaluate for f () = 8 +. + 4 A) B) C) + 8 D) 0 + 8( ) + E) 6 + + 8
PART II: CALCULATOR ALLOWED E6 and E7. The weekly profit that results from selling units of a commodity weekly is given by P = 50-0.00-5000 dollars. E6. Find the actual change in weekly profit that results from increasing production from 5000 units weekly to 505 units weekly. A) $0.00 B) $00.00 C) $99. D) $98.00 E) $8. E7. Use the marginal weekly profit function to estimate the change in weekly profit that results from increasing production from 5000 units weekly to 505 units weekly. A) $0.00 B) $00.00 C) $99. D) $98.00 E) $8. y = E8. Determine whether or not has a relative maimum. If it has a relative maimum, what is the maimum (y value)? A) 0 B) 0.08 C) 0.07 D) 0.079467 E) None eists E9 and E0. Given A = 4 5 7 5 4 8 E9. Find the inverse of A, A -. 44 9 0 0 47 8 A) 64 4 9 B) 0 0 C) 7 50 D) 0 0 78 4 5 7 5 4 8 E) The inverse does not eist. E0. Solve A y z = 5 0 8 67 87 A) 0 B) 65 C) 0 D) 40 E) No solution eists. 50 7 6 5 8
E. Given A = 0 and B =, find A B. 5 0 98 6 8 90 A) B) C) D) E) 5 0 58 8 8 5 5 0 5 E. Find the absolute etrema of f( )= on the closed interval [0, ]. + A) (0, ) and (, ) B) (, 4/5) and (-, -) C) (, 4/5) and (0, 0) D) (, ) and (-, -) E) (, ) and (0, 0) E. Find Lim e ( + ) e 0 A) 0 B) e C) e D) e E) undefined (does not eist) E4. Find the average value of f() = (4-5) on the interval [-, ] (i.e. - # # ). A) -05 B) 05 C) -40 D) 40 E) -85 E5. A sketch of the graphs of f() = 6 - and g() = -5-0 appears on the right. Find the total area of the shaded region. A) 54.75 B).75 C) 49.5 D) 46.75 E) 99.5
ADDITIONAL PROBLEMS PART I: NO CALCULATOR ALLOWED F. Find all of the points at which the graph of f() = 4-4 + 5 has horizontal tangent lines. A) (,0) only B) (,0) only C) (,) and (-,0) D) (,0) and (,) E) (,) only F. Find dy/d if e y + = y y y y e y e A) 0 B) y C) D) E) e e y e y y e y Lim 4 + 4 F. Find. + + A) B) /4 C) -/8 D) 4 E) -4 F4. Find the intervals on which f () = - + is continuous. A) (0, ) (i.e 0 < < ) B) (-4, 4) (i.e. all real ) C) (0, ) (i.e. 0 < < ) D) (0, 4) (i.e. > 0) E) (-4, 4) c (4, 4) (i.e. -4 < < 4 or > 4) F5. f( )= has as a horizontal asymptote + A) y = / B) y = ½ C) y = 0 D) y = E) y = / f ( ) = F6. The slope of the line tangent to at (4, ) is A) 6 B) C) /4 D) -/4 E) F7. The derivative of f() = ( - 5 ) 7 is A) 7( - 5 ) 6 B) 7( - 0) 6 C) 7( - 0)( - 5 ) 6 D) 70( - 5 ) 6 E) ( - 0) 6 F8. The radius r of a circle is increasing at the rate of inches per minute. What is the rate of change in the area when r = 0 inches? A) 6π inch/min B) 400π in /min C) 40π in /min D) 60π in /min E) 0π in /min F9. Find all of the critical numbers for f( )= 9. A) 0 B) -, C) D) -, 0, E) -
F0. Find the derivative of f () = 4 e A) 4 e B) 4 e + 4 e C) 4 + e D) 4 + e E) 4 e f ( ) = F. The function is discontinuous when A) = 0 B) > 0 C) = D) > E) never F. Find ALL of the vertical asymptotes for y = 9 6 A) = ± 4 B) = ± C) y = D) y = 9/6 E) = ±, ±4 5 4 F. Find the derivative of f( )=. + 7 5 0 + 6 A) B) C) D) E) + ( + ) ( + ) ( + ) F4. Find the slope of the line tangent to the graph of f () = / - + 9 at the point (8, ). A) B) -/6 C) D) -4/ E) -8 F5. Determine the coordinate of the point(s), if any, at which the function y = 5+ 6 has a horizontal tangent. A) = 0 B) = -, 5 C) =, -5 D) Such points do not eist. E) = -5, F6. If f () = 8 4-0 + -, find the value of f O() (the second derivative) at =. A) 9 B) 64 C) 9 D) 76 E) 84 4 F7. Find Lim 0 6 A) 0 B) Does not eist (undefined) C) 4 D) -0 E) - F8. If the demand equation is given by p = 9, then the marginal revenue is 0 when A) = 9 B) = C) = 6 D) = E) = 5 5
F9. For what values of is the function f( )= not continuous? A) B) - and C) -5, - and D) - and E) -, and 0 F0. If A =, find A. 4 9 0 6 5 6 6 9 4 6 0 A) B) C) D) E) Undefined F. Perform the matri operations: 0 5 6 + 5 0 7 5 0 0 5 45 0 8 5 4 4 5 8 6 5 9 4 8 4 5 4 5 A) B) C) D) E) 0 5 8 F. Solve the following system of equations: - y = 8 + y = 7 - y = 5 A) (0, ) B) (8, -, -) C) There are infinitely many solutions D) No solution eists E) (5, -) F. Find the following product: [ 5] 8 5 6 4 4 0 0 [ 8 5 6 ] [ 6 5 6] 6 5 6 A) B) C) D) E) 4 0 6 0 0 F4. Find all solutions to the system of equations. + w = 5 y - z = - + y + w = 8 A) (5 - w, - + w, - + w, w) B) (5, -, -, 0) C) (, -, -, 0) D) (4, -, 0, ) E) No solution eists
PART II: CALCULATOR ALLOWED F5. Find Lim + 4 6 A) 0 B) Does not eist (undefined) C) 4 D) -4 E) F6. For what values of is the slope of the line tangent to y = + - + 7 equal to 0? ± A) None B) (-7.785 or 0.04576) C) /7 D) 0 E) ± = ±5.56776 F7. Given f( )= 4 find f O(). + 5+ 7 A) 84 B) 0 C) 59 D) 84/59 E) F8. For which value of does the graph of the function f () = e - ln have a horizontal tangent? Round your answer to the nearest hundredth. A) B).76 C).8 D).9 E) Never f( )= + 4 F9. Find the equation of the straight line tangent to at the point (, -9). y = ( 4 ) + 4 A) None eists (false) B) C) y = - + 57 D) y = - + 4 E) y = -
ANSWERS SAMPLE EXAM A SAMPLE EXAM B SAMPLE EXAM C SAMPLE EXAM D SAMPLE EXAM E ADDITIONAL PROBLEMS PART I PART I PART I PART I PART I PART I A. D A. A A. D A4. D A5. D A6. B A7. A A8. C A9. C A0. D A. C A. B A. A A4. B A5. B A6. C A7. C A8. E A9. D A0. A A. A A. D A. E A4. B A5. A B. B B. A B. A B4. B B5. C B6. E B7. D B8. A B9. D B0. A B. B B. D B. E B4. D B5. D B6. B B7. C B8. E B9. E B0. A B. C B. D B. B B4. A B5. C C. D C. E C. E C4. D C5. B C6. E C7. A C8. D C9. A C0. A C. E C. D C. C C4. C C5. B C6. B C7. B C8. D C9. E C0. A C. A C. C C. C C4. E C5. D D. E D. A D. D D4. A D5. C D6. C D7. A D8. E D9. D D0. C D. B D. B D. C D4. B D5. D D6. A D7. C D8. B D9. C D0. A D. B D. D D. A D4. C D5. B E. B E. D E. A E4. A E5. D E6. D E7. A E8. B E9. E E0. D E. B E. C E. E E4. D E5. C E6. D E7. E E8. D E9. D E0. B E. D E. C E. D E4. C E5. E F. E F. B F. E F4. B F5. C F6. C F7. C F8. E F9. D F0. B F. C F. A F. B F4. D F5. B F6. B F7. E F8. C F9. D F0. B F. D F. D F. C F4. A PART II PART II PART II PART II PART II PART II A6. E A7. B A8. A A9. C A0. B A. D A. D A. D A4. E A5. B B6. A B7. B B8. C B9. A B0. A B. A B. A B. D B4. D B5. C C6. E C7. B C8. C C9. A C0. D C. C C. E C. B C4. E C5. C D6. C D7. D D8. B D9. E D0. D D. B D. E D. E D4. A D5. A E6. C E7. B E8. B E9. A E0. B E. D E. E E. D E4. A E5. A F5. C F6. B F7. D F8. B F9. D