Old Math 220 Exams. David M. McClendon. Department of Mathematics Ferris State University

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1 Old Math 0 Exams David M. McClendon Department of Mathematics Ferris State University Last updated to include exams from Spring 05

2 Contents Contents General information about these exams 4 Exams from 0 to Fall 0 Exam Fall 0 Exam Fall 0 Exam Fall 0 Exam Fall 0 Exam Fall 0 Final Exam Fall 03 Exam Fall 03 Exam Fall 03 Exam Fall 03 Exam Fall 03 Final Exam Spring 04 Exam Spring 04 Exam Spring 04 Exam Spring 04 Exam Spring 04 Final Exam Spring 05 Exam Spring 05 Exam Spring 05 Exam Spring 05 Exam Spring 05 Final Exam Fall 05 Exam Fall 05 Exam Fall 05 Exam

3 Contents.5 Fall 05 Exam Fall 05 Final Exam Exams from 07 to present 3 3. Spring 07 Exam Spring 07 Exam Spring 07 Exam Spring 07 Exam Spring 07 Final Exam

4 Chapter General information about these exams These are the exams I have given between 0 and 05 in Calculus I courses. Each exam is given here, followed by what I believe are the solutions (there may be some number of computational errors or typos in these answers). I have edited these exams to remove questions that do not match the current syllabus of Math 0; so some of them may contain a less than expected number questions. Note that this calculus course has been revised several times over the years, and what was on Exam in past years may not match what is on Exam now. To help give you some guidance on what questions are appropriate, each question on each exam is followed by a section number in parenthesis (like (3.) ). That means that question can be solved using material from that section (or from earlier sections) in my Calculus I Lectures. Last, my exam-writing style has evolved over the years; generally speaking, the more recent the exam, the more likely you are to see something similar on one of your tests. 4

5 Chapter Exams from 0 to 05. Fall 0 Exam. a) (.) In one or two sentences, describe (in your own words) what is meant by the following statement: lim f(x) = L x a b) (.) Let g(x) = x + x. Simplify the expression g(x + h) g(x) g(h).. (.4) Shown below is the graph of some unknown function f: Use the graph to estimate each of the following quantities: 5

6 .. Fall 0 Exam a) f(5) b) lim f(x) x 3 c) lim f(x) x 3 d) lim f(x) x 0 + e) lim f(x) x f) lim f(x) x 5 3. (.4) Sketch the graph of a function f which has all of the following properties: The domain of f is R { 3}; f is discontinuous at 3, and, but is continuous at all other real numbers; lim f(x) = 4; x lim x 3 f(x) = ; lim f(x) = f( ) = 0; x lim f(x) = lim x f(x) = 5. x 4. (3.) Evaluate any six of the following seven limits. x (a) lim x x + 5 (b) lim x x 3 x + 4 x(x 3) (c) lim x 3 + x 3 x (d) lim x 0 x4 sin(x 4 ) (e) lim x 5 x 5 x (f) lim x x + x 3 x 5x + 4 (g) lim x f(x), where x + if x < f(x) = 4 if x = x if x > 6

7 .. Fall 0 Exam Solutions. a) This statement means that if one were to cover up the graph of the function above/below x = a, then the most reasonable guess as to the value of the function at a is L. Equivalently, it means that we would expect to see a hole (perhaps filled in) in the graph at the point (a, L). Equivalently, it means that if we take x values that get closer and closer to a, then the corresponding f(x) values get closer and closer to L. b) By direct calculation:. a) f(5) = 7 g(x + h) g(x) g(h) = (x + h) + (x + h) [x + x] [h + h] = x + xh + h x x h h = xh. b) lim f(x).5 x 3 c) lim f(x) DNE (left- and right-hand limits unequal) x 3 d) lim x 0 + f(x) = e) lim x f(x) = f) lim x 5 f(x) = 3. One possible answer is sketched below: a) This function is continuous except when x = 5, so the limit as x can be found by plugging in x = : x lim x x + 5 = ( ) + 5 = 3 3 =. 7

8 .. Fall 0 Exam b) This is a limit at infinity, so factor out highest power of x: x 3 x + 4 lim x x(x 3) = lim x x 3 x + 4 x 3 3x = lim x x 3 ( x + 4 x 3 ) x 3 ( 3 x ) = lim + 4 x x 3 x 3 x = =. 0 c) Plugging in x = 3 gives 3 = which suggests the answer is ±. Now when x 3 +, x > 3 so the numerator is 3 = > 0 and the denominator is 3 (> 3) which is negative. Thus the limit is =. 0 d) Observe that sin(x 4 ) so x 4 sin(x 4 ) x 4. Now lim x 0 x4 = 0 and lim x 4 = 0, x 0 so by the Squeeze Theorem, lim x 0 x 4 sin(x 4 ) = 0 as well. e) Conjugate the square root, cancel terms and then plug in x = 5: x 5 (x 5)( x + ) lim = lim x 5 x x 5 ( x )( x + ) (x 5)( x + ) = lim x 5 x 4 (x 5)( x + ) = lim x 5 x 5 = lim x + x 5 = 5 + = 4. f) Factor and cancel: x + x 3 lim x x 5x + 4 = lim (x )(x + 3) x (x )(x 4) = lim x + 3 x x 4 = = 4 3. g) Compute left- and right-hand limits: lim f(x) = lim x x (x + ) = ( ) + = ; lim f(x) = lim ) = ( ) =. x + x +(x Since the left- and right-hand limits are both equal to, we have lim f(x) =. x 8

9 .. Fall 0 Exam. Fall 0 Exam. (4.) Use the definition of the derivative (not the rules) to compute the derivative of f(x) = x 3. Hint: One or both of the following algebraic facts may be useful: (a + b) 3 = a 3 + 3a b + 3ab + b 3 a 3 b 3 = (a b)(a + ab + b ). Shown below is the graph of some unknown function f: a) (4.) List all the values of x for which the function f is not differentiable. b) (4.) Estimate f (). c) (4.) Find all values of x for which f (x) = 0. d) (6.4) Let g(x) = (f f)(x). Estimate g (3) (explain your answer). 3. a) (6.) Find the slope of the line tangent to the function f(x) = x x+ when x = 3 (simplify your answer). b) (5.5) Suppose an ant is crawling back and forth along a straight piece of rope, such that its position at any time t 0 is f(t) = 3t 8 t. i. Find the velocity of the ant at time 4 (simplify your answer). ii. Find the acceleration of the ant at time (simplify your answer). 4. (8.4) Air is being blown into a spherical balloon at the rate of 3 cm 3 /sec. How fast is the radius of the balloon changing when the radius of the balloon is 5 cm? Simplify your answer. (You may use the fact that the volume of a balloon of radius r is V = 4 3 πr3.) 5. (6.5) Compute dy dx if x y = 4x + 3y (6.6) Choose any six of the seven parts of this problem: ( ) a) Find d dx 3x 6 3 x. 4 b) Find f (x) if f(x) = 4 sin x sec x

10 .. Fall 0 Exam Solutions c) Find the derivative of f(x) = cot(x 3). d) Find y if y = (9x + 4) 3/. e) Let f(x) = (tan 4 x 7x)(3 cos x + 5x 8 ). Find df dx. f) Find dy dx if y = csc 4x + x 3 x 4. g) Let f(x) = x + x. Find f (x).. There are two solutions, depending on which definition of the derivative that you used: Alternatively, f (x) = lim h 0 f(x + h) f(x) h f (x) = lim t x f(t) f(x) t x (x + h) 3 x 3 = lim h 0 h x 3 + 3x h + 3xh + h 3 x 3 = lim h 0 h h(3x + 3xh + h ) = lim h 0 h = lim(3x + 3xh + h ) h 0 = 3x = 3x. t 3 x 3 = lim t x t x. a) f is not differentiable at x =, or 4. b) f () (the slope of the graph at x = ). c) f (x) = 0 only when x = 0. = lim t x (t x)(t + tx + x ) t x = lim t x t + tx + x = x + x + x = 3x. d) By the Chain Rule, g (3) = f (f(3))f (3) = f (0)f (3) = 0 = a) Find the derivative using the quotient rule: f (x) = x(x + ) (x ) (x + ) = x + x (x + ). Thus the slope of the tangent line is f ( 3) = ( 3) +( 3) ( 3+) =

11 .. Fall 0 Exam b) i. First, differentiate to obtain f (t) = 6t 8 = 6t 4 t t. The velocity at time 4 is therefore f (4) = 6(4) 4 4 = 4 =. ii. From part (i), f (t) = 6t 4 t = 6t 4t /. Differentiate again to obtain f (t) = 6 4 ( /)t 3/ = 6 + t 3/. Thus the acceleration at time is f () = 6 + () 3/ = 6 + = Start by differentiating the volume formula with respect to t; then plug in and solve for dr dt. 5. Differentiate implicitly: 3 00π V = 4 3 πr3 dv dt = 4 dr π3r 3 dt 3 = 4 dr π3(5) 3 dt 3 = 00π dr dt cm/sec = dr dt. d ( x y ) = d (4x + 3y + ) dx dx xy + x y dy dx = 4 + 3dy dx x y dy dx 3dy dx = 4 xy ( x y 3 ) dy dx = 4 xy dy dx = 4 xy x y 3 ( ) d 6. a) dx 3x 6 3 x = d 4 dx (3x6 3x 4 ) = 8x 5 + x 5. b) f (x) = 4 cos x sec x tan x. c) Use the Chain Rule (the inside function is x 3; the outside function is cot x) to obtain f (x) = csc (x 3) 4x. d) Use the Chain Rule (the inside function is 9x + 4; the outside function is x 3/ ) to obtain y = 3 (9x + 4)/ 9 = 7 9x + 4. e) Use the Product Rule to obtain f (x) = [tan 4 x 7x] (3 cos x + 5x 8 ) + [3 cos x + 5x 8 ] (tan 4 x 7x). Now the derivative of tan 4 x is done by the

12 .. Fall 0 Exam Chain Rule (the inside function is tan x and the outside function is x 4 ) and the other derivatives are straightforward. We obtain, as an answer, f (x) = df = [4 dx tan3 x sec x 7](3 cos x+5x 8 )+[ 3 sin x+40x 7 ](tan 4 x 7x). f) First, rewrite y as y = csc 4x + 4 x4/3. Now the derivative of the first term is done by the Chain Rule (the inside function is 4x and the outside function is csc x); the entire derivative is dy = csc 4x cot 4x 4+ dx x/3 = 4 csc 4x cot 4x + 3 x/3. g) Use the Chain Rule (the inside function is x + and the outside function x is x) to obtain f (x) = [ ] x + x = x(x ) x(x +). (x ) x + x x + x

13 .3. Fall 0 Exam 3.3 Fall 0 Exam 3. (9.5) Let f(x) = x x 5. Does the Mean Value Theorem apply to f on the interval [0, 4]? Why or why not? If the Mean Value Theorem does apply, find all values of c which solve the equation in the Mean Value Theorem.. (7.) Find the global maximum value, and the location of the global maximum value, of the function f(x) = + sin 3x on the interval [ 0, π ]. 3. (7.) Shown below is the graph of some unknown function f: a) List all the critical points of f. b) Find the absolute minimum value of f on [ 3, 7]. c) Give the x coordinates of any local maxima of f. d) List the (approximate) x values of all inflection points of f. 4. (7.4) Farmer Bagwell wants to build a rectangular pen against his house and barn. In particular, he wants to put up a fence along the dashed lines shown in the figure on the next page. What dimensions should Farmer Bagwell build his pen, if he wants the area of the pen to be as large as possible, but if he only has 500 feet of fence available? 3

14 .3. Fall 0 Exam 3 60 ft House 40 ft Barn 5. (8.3) Use Newton s Method with two iterations and initial guess to approximate 3, the cube root of. Write your answer as a fraction in lowest terms. 6. (8.) The range R of a certain projectile shot at angle θ from the ground is given by R(θ) = 00 sin θ. Use quadratic approximation to estimate the range of this projectile when the angle at which it is shot is π/4 +.0 radians. 4

15 .3. Fall 0 Exam 3 Solutions. Since f is cts on [0, 4] and diffble on (0, 4) with derivative f (x) = x, the MVT applies. The equation we need to solve for c is f (c) = f(b) f(a) ; b a the left-hand side is c and the right-hand side is f(4) f(0) 4 0 So we set c = to obtain c =. = 3 ( 5) 4 =.. Start by differentiating f, using the Chain Rule, to obtain f (x) = 3 cos 3x. This derivative exists everywhere, so we find critical points by setting f (x) = 0. This gives us 3 cos 3x = 0, i.e. cos 3x = 0. Thus 3x is an angle with cosine equal to zero, so 3x = π or 3x = 3π and consequently our critical points are x = π and x = π. Plug the critical points and endpoints into f to find the 6 maximum value: x = 0 f(0) = + sin 3(0) = x = π 6 x = π f(π/6) = + sin(3π/6) = 3 f(π/) = + sin(3π/) = So the maximum value is 3, occuring at x = π a) f (x) = 0 where the tangent line is horizontal; this is at x = 7, x = 3 and x = 3. f (x) DNE where the graph has a cusp; this is at x =. So the critical points are x = 7, 3,, 3. b) This is the minimum y value on [ 3, 7] which is. c) x = 7, x = 3. d) x = 3, x = First, let x and y be the side lengths indicated on the picture below: 5

16 .3. Fall 0 Exam 3 y x 60 ft House 40 ft Barn Now the total amount of fence used is x+y+(x 40)+(x 60) = x+y 00. This should be equal to 500, so we have x+y 00 = 500, i.e. x+y = 600, i.e. x + y = 300, and this implies y = 300 x. The goal of this problem is to maximize the area of the rectangle, so we write an equation for the area: A = xy. Now from the earlier equation y = 300 x, we can plug in this expression for y in our area expression and obtain A(x) = x(300 x) = 300x x. We seek to maximize this for x [40, 40]. To do this, differentiate A to obtain A (x) = 300 x. Solving A (x) = 0 gives x = 50 as our only critical point. Now we test the critical points and endpoints in the equation for the area: x = 50 A = x(300 x) = 50 = 500 x = 40 A = x(300 x) = 40(300 40) = = 0400 x = 40 A = x(300 x) = 40(300 40) = = 4400 The maximum area is found when x = 50; the dimensions of the pen are x by y which is 50 by To approximate x = 3 using Newton s method, first find a function f such that f(x) = 0. Now if x = 3, then x 3 = and x 3 = 0 so f(x) = x 3. We apply Newton s method to this function (notice f (x) = 3x ) with initial guess x 0 = : x = x 0 f(x 0) f (x 0 ) = 3 3 = 3 = 4 3 x = x f(x ) f (x ) = 4 3 (4/3)3 3 (4/3) = = = = =

17 .3. Fall 0 Exam 3 6. If R(θ) = 00 sin θ then R (θ) = 400 cos θ and R (θ) = 800 sin θ. Therefore R(π/4) = 00 sin π = 00; R (π/4) = 400 cos π = 0 and R (π/4) = 800 sin π = 800. Thus the quadratic approximation to R when a = π/4 is Q(θ) = Q(π/4) + Q (π/4)(x π/4) + Q (π/4)(x π/4) = (x π/4) + ( 800)(x π/4) = (x π/4). To solve the problem, plug in θ = π/4 +.0 to get Q(θ) = (.0) = =

18 .4. Fall 0 Exam 4.4 Fall 0 Exam 4. a) (9.5) Precisely state one part of the Fundamental Theorem of Calculus. b) (9.5) Precisely state the part of the Fundamental Theorem of Calculus you did not state in part (a).. a) (.3) Find all solutions to the ODE y = xy. b) (.) Shown below is the vector field associated to some first-order ODE y = g(x, y): i. On the above vector field, sketch the graph of the solution to this ODE passing through the point (, ). ii. Let y = f(x) be the solution to this ODE passing through (0, 0). Find lim x f(x). 3. (9.) Consider the function f(x) = x. Estimate the integral 7 x+ 0 f(x) dx by computing the right-hand Riemann sum for f associated to the partition P = {0,, 3, 7}. Simplify your answer. 4. (0.3) Evaluate any six of the following seven integrals. a) 3x(x + ) dx b) (4 csc x + cos x) dx c) (x x) dx d) 4x sin(3x + ) dx e) 3 6 x dx f) 0 4x (3x + ) 3 dx g) π/4 0 cos 3x dx 8

19 .4. Fall 0 Exam 4 Solutions. a) Let f be continuous on [a, b] and define F (x) = x a f(t) dt. Then F is differentiable on [a, b] and F (x) = f(x). b) Suppose f is continuous on [a, b] and let F be any antiderivative of f. Then b a f(x) dx = F (b) F (a).. a) Separate variables and integrate both sides: dy dx = xy y dy = x dx y = x + C. b) i. Not shown here. The graph looks something like a bell curve which goes to 4 as x approaches either or. ii. The graph of this solution moves down to height 4 as it moves to the extreme right, so this limit is Given our partition, we have x = ; x = ; x 3 = 4. Since we are doing a right sum, we have c =, c = 3, c 3 = 7. So the Riemann sum is 3 j= f(c j ) x j = f() + f(3) + f(7)4 = () () + 7 (4) = a) 3x(x + ) dx = (3x + 3x) dx = x + 3 x + C. b) (4 csc x + cos x) dx = 4 cot x + sin x + C. c) (x x) dx = x x x3/ 3/ + C = x x 4 3 x3/ + C. d) Let u = 3x + so that du = 6x and du = 6xdx. Therefore du = 4xdx and dx 3 substituting this into the integral, we get sin u du = cos u + C = cos(3x + ) + C. e) 3 6 x dx = 3 6x dx = 6x 3 = [ 6(3) ] [ 6() ] = = + 6 = 4. f) Let u = 3x + so that du = 6x and du = 6x dx so 4du = 4x dx. Also, dx when x = 0, u = 3(0) + = and when x =, u = 3() + = 4. Putting this all together, we have 4 4u3 du = u 4 4 = = 55. g) Set u = 3x so that du = 3 and therefore du = 3dx, so du = dx. When dx 3 x = 0, u = 3 0 = 0 and when x = π, u = 3 π. Substituting this back into 4 4 the integral, we have 3π/4 0 cos u du = sin u 3π/ = sin ( ) 3π 3 4 sin 0 = 3 3 ( ) = 6. 9

20 .5. Fall 0 Exam 5.5 Fall 0 Exam 5. (6.5) Let f(x) = 3e x + e x. Find (f ) (4).. (8.) Evaluate each limit: (a) lim (4 arctan x) x (b) lim x e 6x (c) lim x 3 ln x 3. (6.6) Find the derivative of each of the following functions: a) y = 0e 4x b) f(x) = ln x + arctan 8x c) y = (x ) 3x 4. (0.3) Evaluate each integral: (a) 6x x dx (b) e 3x dx (c) (x ) dx x 5. a) (.4) Suppose that the value of an antique grows exponentially. If the antique was worth $50 in 930 and was worth $600 in 000, what will it be worth in 030? Hint: let t = 0 correspond to the year 930. b) (.4) Suppose that the percentage P of American families owning a DVD player t years after the year 990 is modeled by the following differential equation: dp dt = y(80 y) 40 Suppose that in 990, 0% of families owned a DVD player. i. What percentage of American families will own a DVD player in 05? ii. What percentage of American families will own a DVD player in the long run (i.e. as t )? 0

21 .5. Fall 0 Exam 5 Solutions. Observe that f (x) = 3e x + e x > 0, so f is strictly increasing, hence. Notice that f(0) = 3e 0 + e 0 = 3 + = 4, so by the Inverse Function Theorem, (f ) (4) = = =. f (0) 3e 0 +e 0 5 π. a) lim (4 arctan x) = 4 = π. x b) lim x e 6x = e = 0. c) lim x 3 ln x =. 3. a) By the Chain Rule, y = 0e 4x 4 = 40e 4x. b) Use the Chain Rule on the second part of this function to obtainf (x) = + 8 = + 6. x +(8x) x +64x c) Take the natural logarithm of both sides and rewrite the right-hand side using a log rule; then differentiate both sides implicitly: y = (x ) 3x ln y = ln [ (x ) 3x] ln y = 3x ln(x ) y y = 3 ln(x ) + x x 3x [ y = y 3 ln(x ) + x ] x 3x [ ] y = (x ) 3x 3 ln(x ) + 6x x 4. a) Let u = x so that du = 3x dx and du = 6x dx. Therefore 6x x dx = u du = ln u + C = ln(x3 + 5) + C. b) Let u = 3x so that du = 3 dx and du = dx. Therefore 3 c) Multiply out: e 3x dx = 3 eu du = 3 eu + C = 3 e3x + C. (x ) x x + dx = dx = x x ( x + ) x dx = x x+ln x+c.

22 .5. Fall 0 Exam 5 5. a) We have y = Ce kt ; the initial value is C = 50. When t = 70, y = 600, so we have 600 = 50e 70k. Solve for k by dividing both sides by 50, taking natural logs, and finally dividing by 70. This yields k = ln. 70 Now, the year 030 corresponds to t = 00, so at that time y = 50e k(00) = 50e ln = 50e (0/7) ln = 50 0/7. b) i. We are given a differential equation of the form dp dt = ky(l y) which is a logistic equation whose solution is P = L + Ce klt. We are given k = and L = 80, so our equation is therefore 40 P = 80 + Ce t. Now when t = 0, P = 0 so we can solve for C: 0 = 80 + Ce = 8 + C + C = 8 = Now the year 05 corresponds to t = 5, so at this time P = e = e. 50 ii. This is the carrying capacity L, which is 80 percent.

23 .6. Fall 0 Final Exam.6 Fall 0 Final Exam. (3., 8.) Compute any five of the following six limits. x x 6 (a) lim x 3 x 7x + x + 3 (b) lim x + x (c) lim f(x), where x { 3x if x f(x) = 5 if x = e 3x (d) lim x 0 x x (e) lim x 4 x 4 3x + (f) lim x x 4 5x + 3. (6.6) Choose any five of the following six problems: a) If f(x) = x cos x, find f (x). b) If y = arctan x x 3/, find y. c) Find dy dx if x y 6e x = 4. d) Find f (x) if f(x) = ln(sin x). e) Find f ( ) if f(x) = (3x 4x + ) 0. f) Find h (x) if h(x) = 3 x + cot(3x) sec x. 3. (0.3) Evaluate any five of the following six integrals: (a) (6x 7 + x) dx (b) (c) π 0 3 sin x dx 5 x dx (d) 3 e4x dx ( (e) x ) dx x (f) e x sin(e x ) dx 3

24 .6. Fall 0 Final Exam 4. Classify the following statements as true or false: a) (7.) If f (x) = 0, then x is a critical point of f. b) (7.) If f (x) = 0, then x is the coordinate of a local extrema of f. c) (7.) If f (x) = 0, then x is the location of an absolute extrema of f. d) (9.5) Two antiderivatives of the same function must differ by a constant. e) (7.3) If f (x) = 0, then x is an inflection point of f. f) (4.) If f is differentiable at x = c, then f is continuous at x = c. 5. (.-.4, 4., 7., 7.3, 9.) The graph of some unknown function f is given below: Use the graph to estimate the following: a) lim x + f(x) b) x lim f(x) c) lim f(x) x d) f ( 5) e) a number x such that f is continuous but not differentiable at x f) the x coordinate of a local minimum of f g) all intervals on which f is concave up h) 5 7 f(x) dx

25 .6. Fall 0 Final Exam 6. a) (7.3) Find the x coordinates of all local extrema of the function f(x) = x 3 + 6x + 36x + ; classify these extrema as local maxima or local minima. b) (6.4) Find the equation of the line tangent to the function h(x) = x 3 where x = Choose one of the following two problems: a) (7.4) A rectangle has one side on the x axis, one side on the y axis, and one corner in the first quadrant on the graph of y = e x (see the figure below). Find the maximum area of such a rectangle. b) (8.4) Suppose a ladder of length 5 ft leans up against a wall. If the top of the ladder is sliding down the wall at a rate of ft/sec, how fast is the bottom of the ladder sliding away from the wall at the instant the top of the ladder is feet off the ground? 5

26 .6. Fall 0 Final Exam Solutions. a) lim x 3 x x 6 =lim x 7x+ x 3 (x 3)(x+) (x 3)(x 4) =lim x 3 x+ = 5 = 5. x 4 x+3 b) lim = 5 =. x + x +0 c) Disregard the special rule for the function at x = ; the limit is lim 3x = x 6. d) lim x 0 e 3x x = 0 0 e) lim x 4 x x 4 = 0 0 f) lim x 3x + x 4 5x +3 = lim x L 3e =lim 3x = 3. x 0 L = lim x x 4 =. 4 x 4 ( 3 x + ) x x 4 ( 5 x + 3 x x 4 ) = lim 3 x + x = x + 3 x = 0.. a) By the product rule, f (x) = cos x x sin x; and by the product rule again, f (x) = sin x [ sin x + x cos x] = 4 sin x x cos x. b) By the quotient rule, f (x) = x + x3/ 3 x/ arctan x x 3. c) Differentiate implicitly to get xy + x dy dx 6ex = 0; solve for dy to get dx dy = 6ex xy. dx x d) By the Chain Rule, f (x) = cos x = cot x. sin x e) By the Chain Rule, f (x) = 0(3x 4x + ) 9 (6x 4) so f ( ) = 0( ) 9 ( 0) = f) Use the Chain Rule on the middle term to get h (x) = 3 x /3 3 csc (3x) sec x tan x. 3. a) (6x 7 + x) dx = 6 x8 + x3/ C = x x3/ + C. b) π 0 3 sin x dx = 3 cos x π 0 = 3 cos π + 3 cos 0 = = 6. c) 5 x dx = 5 arcsin x + C. d) Let u = 4x so du = 4, du = 4dx and finally du = dx. Thus dx 4 3 e4x dx = 3 4 eu du = 6 eu + C = 6 e4x + C. e) ( x x ) dx = ( x x /) dx = ln x x + C. f) Let u = e x so that du dx = ex and du = e x dx. Then e x sin(e x ) dx = sin u du = cos u + C = cos(e x ) + C. 4. a) TRUE by definition of critical point. b) FALSE; the sign of the derivative may not change at x. c) FALSE; same reason as (b). d) TRUE; this is a theorem from class called the Antiderivative Theorem. 6

27 .6. Fall 0 Final Exam e) FALSE; the sign of f may not change at x. f) TRUE; this is a theorem from class called Differentiability implies continuity. 5. a) This is the limit as x approaches from the right, which is. b) This is the height of the horizontal asymptote of f which is. c) This is the limit as x approaches from the left, which is. d) This is the slope of the graph at x = 5, which is. e) This asks for x values where the graph has a cusp or corner, i.e. x = 7. f) x = 0 and x = are local minima. g) (, ) is the only interval on which f is concave up. h) This is the area under the graph from x = 7 to x = 5; this area is the area of a triangle with height and base so it is bh = =. 6. a) f (x) = 3x + x + 36 = 3(x 6)(x + ); setting f (x) = 0 we find critical points x = 6, x = (there are no x where f (x) DNE). We classify these CPs here by computing the second derivatives at these points: f (x) = 6x + so f (6) = 4 < 0 so 6 is a local maximum; f ( ) = 4 > 0 so is a local minimum. b) First, h(6) = =. Now h (x) = (x 3) = so h (6) = (x 3) =. Finally, by the point-slope formula, the tangent line has (6 3) 9 equation y = (x 6) a) If we let x be the x coordinate of the northeast vertex of the rectangle, then the width of the rectangle is x and the height is e x, so the area of the rectangle is A(x) = xe x. We seek to maximize A(x) on [0, ). First, we find critical points: by the product rule, A (x) = e x xe x = e x ( x). Solving A (x) = 0, we can cancel the e x term (since it is never zero) and obtain x =. Now, test the critical points and endpoints in the rule for A: for the critical point x =, we have A() = e = and for e the endpoint x = 0, we have A(0) = 0 so the maximum area is. e b) Let x be the distance from the bottom of the ladder to the wall, and let y be the distance from the top of the ladder to the ground. Then x + y = 5. We are given y = and dy = and asked to solve for dt. Differentiating the above equation with respect to t, we obtain dx dt x dx dy + y dt dt = 0. When y =, x = 9 (we get this by solving x + = 5 for x). Plugging all the known values into the above equation yields 9 dx + ( ) = dt 0 so dx = 48 = 8 ft/sec. dt 8 3 7

28 .7. Fall 03 Exam.7 Fall 03 Exam. a) Suppose you want to define the function f(x) = x x+ in Mathematica. Write the code you would use to define f. b) Suppose you want to produce a graph of the function f(x) = 3x +, where the range of x values shown on the graph is from 4 to 5. Write the code you would use in Mathematica to produce this graph.. (.4) Shown below is the graph of some unknown function f. Use the graph to estimate the following (if the quantity does not exist or is ±, say so): a) f() b) lim f(x) x c) lim f(x) x 0 d) lim f(x) x 4 e) lim f(x) x 4 + f) lim f(x) x 6 g) lim f(x) x h) lim f(x) x i) the equation(s) of all vertical asymptote(s) of f 3. a) (.3) Evaluate lim ln x. x

29 .7. Fall 03 Exam b) (.3) The answer to the limit in part (a) reveals some particular characteristic about the graph of f(x) = ln x. What characteristic does it reveal? c) (3.) Evaluate lim x x 4x x +5x+6. d) (3.) Let g(x) = x 4x (the function from the limit in part (c)). The x +4x+4 function g is discontinuous at x =. Why? e) (3.) Based on your answer to part (c), what kind of discontinuity must the function g have at x =? f) (.4, 3.) Evaluate lim x e 3x+. g) (.4) The answer to the limit in part (e) reveals some particular characteristic about the graph of h(x) = e 3x+. What characteristic does it reveal? 4. (3.) Evaluate any five of the following six limits. (a) lim x + x 6 4 x (b) lim x + x x+ (c) lim x 4x + 3x 3x x 4 x (d) lim x x ( ) x (e) lim cos x π 4 (f) lim f(x), where x 3 3x if x < 3 f(x) = 4 if x = 3 x x + if x > 3 9

30 .7. Fall 03 Exam Solutions. a) f[x_] = x^ - x + b) Plot[3x +, {x, -4, 5}]. a) 4 b) c) 3 d) e) DNE (too many wiggles) f) DNE (left- and right-hand limits unequal) g) - h) DNE (no HA to the right) i) x = 4, x = 4 3. a) b) f has a vertical asymptote x = 0. c) 8 d) Because g( ) DNE. e) Removable (since lim x g(x) exists). f) 0 g) h has horizontal asymptote y = a) Start by plugging in x = to get, which is ±. To determine the 0 sign, notice that x + means x >, so x > 4, so 4 x < 0 so the denominator is negative. Since the numerator is also negative, the whole expression is positive so the answer is. b) Plug in x = : lim x + x = x = = =. + c) This is a limit at infinity, so factor out the highest power of x that occurs: 4x + 3x lim x 3x x = lim x x (4 + 3 x x ) x ( 3 x ) = lim x =. d) Plug in x = 8 to get = 0. Thus, we need to conjugate the square 0 30

31 .7. Fall 03 Exam roots: 4 x lim x x = lim (4 x)(4 + x) x 8 + (8 x)(4 + x) 6 x = lim x 8 + (8 x)(4 + x) (8 x) = lim x 8 + (8 x)(4 + x) = lim x x = = 4. e) lim x π cos ( ) x 4 = cos( π ) =. 4 f) Compute left- and right-hand limits. When x < 3, we have lim f(x) = 3(3) = 7 x 3 and when x > 3, we have lim f(x) = x = 8. Since these + one-sided limits are unequal, lim f(x) DNE. x 3 3

32 .8. Fall 03 Exam.8 Fall 03 Exam. a) Suppose you want to find the derivative of the function f(x) = 3e x sin x when x = 0. Write Mathematica code you could use to solve this problem. b) If you typed the following code into Mathematica: D[/x+, {x, }] what would the output be?. a) (4.) On the axes below, sketch the graph of a function f which is continuous at all x, but is not differentiable at x = and x = b) (4.) Given the graph of the function h below, estimate h (5) c) (5.5) Suppose g is some function which satisfies g( 4) =, g ( 4) = /3, and g ( 4) = 4. Sketch a picture of what the graph of g looks like near x =

33 .8. Fall 03 Exam 3. Suppose someone is using the definition of derivative to compute some derivative, and after writing out this definition obtains the following limit: 3(π + h) sin(π + h) 3π sin π lim h 0 h a) (4.) What function is the person trying to differentiate? b) (4.) At what value of x is the function trying to differentiate the function? c) (6.) What is the value of this limit? 4. Suppose an ant is crawling back and forth along an axis so that its position at time t is f(t) = t 3 t + 5t. a) (5.) Find the velocity of the ant at time. b) (5.5) At the instant when t =, is the ant speeding up, or slowing down? Explain. 5. (6.6) Answer any five of the following six problems. a) Find f (x) if f(x) = 3 csc x + tan x 5e x. b) Find the slope of the line tangent to the function f(x) = x 3 x 4 +5x when x =. c) Find dy dx if y = 4 arcsin x xex. d) Suppose x 3 + xy = 4. Find dy when x = and y = 3. dx e) Find the derivative of f(x) = (x 3 + ln x)(3x 4 x). f) Find g ( π ) if g(x) = ln(sin x). 3 33

34 .8. Fall 03 Exam Solutions. a) f[x_] = 3 E^x Sin[x] f [0] b) This is asking for the second derivative of f(x) = x +. Since f (x) = x = x, f (x) = x 3. Thus the output would be x 3.. a) Here is one such function: b) h (5) is the slope of this function at x = 5, which is about. c) The function should pass through ( 4, ), have slope at that point (so 3 go up slightly from left to right), and lie below its tangent line since g ( 4) < 0. So it should look something like a) f(x) = x sin x. b) x = π c) The value is f (π). Since f(x) = x sin x, by the Product Rule f (x) = sin x + x cos x so f (π) = sin π + π cos π = 0 + π = π. 4. a) v(t) = f (t) = 3t 4t+5; at time the velocity is v() = 3() 4()+5 = = 4. b) The acceleration at time t is a(t) = v (t) = 6t 4. At time, a() = 6() 4 =. Since the acceleration at time is positive, the ant is speeding up. 5. a) f (x) = 3 csc x cot x + sec x 5e x. b) By the Quotient Rule, f (x) = (4x)(x4 +5x) (4x 3 +5)(x 3) (x 4 +5x). So f () = (4)(+5) (4+5)( 3) (+5) = 4 ( 9) 36 = =. 34

35 .8. Fall 03 Exam c) dy dx = 4 x [e x + xe x ] = 4 x e x xe x. The second term requires the Product Rule, then the Chain Rule. d) Differentiate implicitly: d dx x 3 + xy = 4 ( x 3 + xy ) = d dx (4) 3x + y + x dy dx = 0 Now plug in x =, y = 3 to get 3() dy dx dy = 6. dx e) By the Product Rule, f (x) = = 0. Solve for dy dx ( 3x 4 + ) (3x 4 ( x) + x 3 ) (x x 3 + ln x ). x to get f) By the Chain Rule, g (x) = cos x = cot x. Thus sin x g ( π) = cot ( ) π 3 3 = 3. 35

36 .9. Fall 03 Exam 3.9 Fall 03 Exam 3. a) (8.) Suppose f is some unknown function which satisfies f() = 4 f () = f () = 6. Find the quadratic approximation to f when a =. b) (8.) Use linear approximation to estimate e /4.. (8.3) Use Newton s method, with initial guess and one step, to approximate a solution of the equation x = Classify the following statements as TRUE or FALSE. a) (7.) Critical points of a function f are values x for which f (x) = 0. b) (7.3) If a function f is concave up, then the function lies below its tangent lines. c) (7.3) If f is decreasing, then f is concave up. d) (8.) 0 is an indeterminate form. e) (7.) The absolute maximum of a function must also be a local maximum. 4. (8.) Evaluate the following limits: (a) lim x 0 sin x x tan x (b) lim x 4 x x 6 5. a) (7.) Find all critical point(s) of the function f(x) = x 4 x 3 +. b) (7.3) Determine whether the function f(x) = xe x is increasing or decreasing on the interval (0, ). Justify your answer. 6/7. (7.4) Solve two of these three word problems: a) A box of volume 4 cubic units is to be constructed with no top and a square base. If the box is to be made with the smallest possible amount of material, what should be the height of the box? b) A rectangle in the xy plane is situated so that its bottom is on the x axis, its left side is on the y axis, and its upper-right corner is on the graph of the function f(x) = 3 x 3. Find the largest possible area of such a rectangle. c) A farmer wants to build a corral in the shape of the thick lines in the picture on the next page (he does not need to build a fence along the river). 36

37 .9. Fall 03 Exam 3 RIVER If the farmer has 40 feet of fence available, what is the largest area his corral can enclose? 37

38 .9. Fall 03 Exam 3 Solutions. a) Q(x) = f(a) + f (a)(x a) + f (a)(x a) = 4 (x ) + 3(x ) 3. b) Let f(x) = e x and a = 0. Then e /4 = f(/4) L(/4) = f(0) + f (0)(/4 0) = + (/4 0) = Set one side of the equation x = 3 equal to zero to obtain x 3 = 0. So f(x) = x 3 and f (x) = x. If x 0 =, then x = x 0 f(x 0) f (x 0 ) = 3 = 4 = a) This is FALSE (you also have to consider where f (x) DNE). b) This is FALSE (if f is concave up, then the function lies above its tangent lines). c) This is FALSE (if f is decreasing, then f < 0 so f is concave down). d) This is FALSE ( 0 = ). e) This is TRUE (by definitions of absolute and local maximum). 4. Use L Hôpital s Rule on both limits: a) b) sin x x lim x 0 tan x = 0 cos x = lim 0 x 0 sec x = x 0 lim x 4 x 6 = = lim 0 x 4 x x = 4 8 = 3. = a) First, compute the derivative: f (x) = 4x 3 36x. Set this equal to zero and solve for x (there are no x where f DNE). After factoring, you get 4x(x 3)(x + 3) = 0 so the critical points are 0, 3 and 3. b) By the Product Rule, the derivative is e x + xe x ( ) = e x +xe x = e x ( + x). When x (0, ), e x > 0 and + x < 0 so f (x) < 0 so f is decreasing on (0, ). 6/7. a) Let x be the lengths of the sides of the base and let h be the height of the box. Since the volume is 4, we have 4 = x h, and by solving this for h we see h = 4 x. Now the amount of material used is the surface area of the box, and since the box has no top this surface area is S = x + 4xh. 38

39 .9. Fall 03 Exam 3 Substituting in h = 4 x for h in this equation, we obtain the utility S(x) = x + 4x 4 x = x + 6 x. Our goal is to minimize this function on (0, ). To do this, take the derivative: S (x) = x 6 x = x3 6. x S (x) DNE when x = 0; now set S (x) = 0 (i.e. set the numerator of S equal to zero) and solve for x: x 3 6 = 0 x 3 = 6 x 3 = 8 x = So the critical points are x = 0, x =. Since S < 0 when x < and S > 0 when x >, the function goes from decreasing to increasing at x =, i.e. x = is the location of the minimum value of S. The question asks for h; h = 4 x = 4 = unit. b) The area of this rectangle is A(x) = x(3 x 3 ). We have to maximize this area on the interval of x values [0, 3 3]. First, take the derivative and set it equal to zero (there are no x such that A (x) DNE). A(x) = 3x x 4 so A (x) = 3 4x 3. Now 3 4x 3 = 0 3 = 4x 3 8 = x 3 = x Now test the critical point and the endpoints: CP x = : A() = (3 8) = 48 EP x = 0 : A(0) = 0(3 0) = 0 EP x = 3 3 : A( 3 3) = 3 3(3 3) = 0 Therefore the maximum area is 48 square units. c) Let y be the distance all the way across the bottom of the corral, and let x be the vertical distance of each of the fences touching the river. Since 39

40 .9. Fall 03 Exam 3 the farmer has 40 feet of fence, we have y + 4x = 40, i.e. y = 40 4x. Now the total area of the corral is A = xy = x(40 4x) = 40x 4x ; our goal is to maximize this area on the interval [0, 60]. Take the derivative and set it equal to zero: A (x) = 40 8x; 0 = 40 8x when x = 30 so x = 30 is the critical point. Now test the critical point and endpoints: CP x = 30 : A(0) = 30( ) = 30(0) = 3600 EP x = 0 : A(0) = 0(40 0) = 0 EP x = 60 : A(60) = 60( ) = 60(0) = 0 Therefore the maximum possible area is 3600 square feet. 40

41 .0. Fall 03 Exam 4.0 Fall 03 Exam 4. Let f(x) = x and let P = {0,, 4, 9}. a) (9.) Find P. b) (9.) Find the right-hand Riemann sum for f associated to P. c) (9.) Is the Riemann sum you found in part (b) an overestimate, or an underestimate of the actual area under f from 0 to 9? Explain.. (0.) Evaluate each of the following four integrals: (a) (3 sin x + csc x ) dx (b) ( e x + 4 ) dx x (c) ( 6x ) dx (d) ( 7x ) dx x 3. (0.3) Evaluate any three of the following four integrals. (c) (a) (b) (d) (x + ) dx x e 4x dx x cos(x 3 + 5) dx 0 0x 3 x 4 + dx 4. a) (.3) Solve the following differential equation (solve your answer for y in terms of x): y = xy b) (.3) Solve the following initial value problem: { y = x 4 + x 4 y y() = 0 4

42 .0. Fall 03 Exam 4 5. (.) Here is a picture of the vector field associated to some first-order ODE y = g(x, y): a) Give the explicit equation of one solution to this ODE. b) Let y = h(x) be the solution to this ODE passing through ( 4, 0). i. Find lim x h(x). ii. lim x h(x). c) Let y = k(x) be the solution to this ODE passing through (0, ). Estimate k(5). d) Let y = f(x) be the solution to this ODE passing through ( 3, 8). Sketch the graph of y = f(x) on the vector field above (if you have drawn other things on the picture above, label the graph you want graded here with f(x) ). 6. (.4) Suppose that the number of bunnies in a field grows exponentially. If initially, there are 0 bunnies, and if three months later there are 00 bunnies, how many bunnies will be in the field after ten months? (Your answer should be written so that it does not contain e or ln.) 4

43 .0. Fall 03 Exam 4 Solutions. Let f(x) = x and let P = {0,, 4, 9}. a) P = 5, the largest width of any subinterval. b) For the right-hand Riemann sum, c =, c = 4 and c 3 = 9. So the Riemann sum is f f(c k ) x k = f() + f(4) 3 + f(9) 5 = = 44. k= c) Since f is increasing, the right-hand sum is the upper sum, which overestimates the area under f from 0 to 9.. a) (3 sin x + csc x) dx = 3 cos x cot x + C. b) ( e x + 4 x) dx = e x + x + C. c) (6x ) dx = [x 3 x] = ( 3 ) ( ) = 4 = a) (x+) x dx b) Let u = 4x so that du = 4 and du = dx. After substituting, we get dx 4 4 eu du = 7 eu + C = 7 e4x + C. c) Let u = x so that du = dx 4x and therefore 3du = x dx. After substituting, we get 3 cos u du = 3 sin u + C = 3 sin(x 3 + 5) + C. d) Let u = x 4 + so that du = dx 4x3. Thus 5du = 0x 3 dx so after substituting (don t forget to change the limits) we get 5 du = [5 ln u u] = 5 ln. 4. a) Rewrite the derivative in Leibniz notation: = dx xy. Then separate the variables by multiplying through by dx and y to get y dy = x dx. Integrate both sides to get the solutions 3 y3 = x + C. Solve for y by multiplying through by 3 and taking a cube root of both sides to get dy y = 3 3 x + 3C. b) As in part (a), rewrite the derivative in Leibniz notation and separate 43

44 .0. Fall 03 Exam 4 the variables: y = x 4 + x 4 y dy dx = x4 ( + y ) + y dy = x4 dx + y dy = x 4 dx arctan y = 5 x5 + C Now plug in the initial condition y() = 0 to solve for C: arctan 0 = C C = 5 Thus the particular solution is arctan y = 5 x a) y = 5 and y = 3 are solutions. b) i. As x goes to the left, the graph of h(x) heads toward y = 5 so the answer is 5. ii. As x goes to the right, the graph of h(x) eventually bends downward and heads toward y = 3, so the answer is 3. c) k(5). d) The graph is shown here: Since the growth is exponential, the equation governing the number y of bunnies at time t is y = Ce kt. We are given that when t = 0, y = 0 so C = 0. We are also given that when t = 3, y = 00, so we have 00 = 0e k 3 Solving for k, we get e 3k = 5 so 3k = ln 5 so k = ln 5. Now after ten months, 3 we set t = 0 and solve for y: y = 0e 3 ln 5 0 = 0e 0 3 ln 5 = 0 5 0/3. 44

45 .. Fall 03 Final Exam. Fall 03 Final Exam. (3., 8.) Compute any five of the following six limits. x 3 x + 4 (a) lim x x + 4 x + (b) lim x 5 5 x (c) lim x 4 x x 4 x + 7x + (d) lim x 0 cos(4x) x (e) lim x 3 x 3 x (f) lim x 7e 4x. (6.6) Solve any five of the following six problems: a) If f(x) = x 3 sin x, find f (x). b) If y = 3x x, find cot x y. c) Find dy if y x cos y = 5. dx d) Find f (x) if f(x) = ln(x 4 + ln x). e) Find f (x) if f(x) = ex 3x tan x 7. f) Find h () if h(x) = (x +) 3 x. 3. (0.3) Evaluate any five of the following six integrals: ( (a) 4x + 3 ) dx x x (b) π/ π/4 (4 cos x + sin x) dx (c) 0e 4x dx (d) 3 (x ) dx (e) 6x 3 cos(x 4 3) dx (f) 0 x x 3 + dx 45

46 .. Fall 03 Final Exam 4. (.-.4, , 9.) The graph of some unknown function f is given below: Use the graph to estimate the following: a) lim x 4 f(x) b) lim x f(x) c) x lim f(x) d) lim f(x) x e) lim f(x) x f) f() g) lim x 7 f(x) h) 4 f(x) dx i) the x coordinate of a local minimum of f j) the x coordinate of an inflection point of f 46

47 .. Fall 03 Final Exam 5. (.,., 4., 5.5, 9.4) The graph of some unknown function f is given below: Based on looking at the graph, for each of the following quantities, circle POSITIVE if the quantity is positive, circle NEGATIVE if the quantity is negative, circle ZERO if the quantity is zero, and circle DNE if the quantity does not exist. (a) f(3) POSITIVE NEGATIVE ZERO DNE (b) f ( ) POSITIVE NEGATIVE ZERO DNE (c) f (6) POSITIVE NEGATIVE ZERO DNE (d) f ( 7) POSITIVE NEGATIVE ZERO DNE (e) f ( 5) POSITIVE NEGATIVE ZERO DNE (f) f (3) POSITIVE NEGATIVE ZERO DNE (g) 3 4 f(x) dx POSITIVE NEGATIVE ZERO DNE (h) lim f(x) POSITIVE NEGATIVE ZERO DNE x 47

48 .. Fall 03 Final Exam 6. Throughout this problem, f is the function f(x) = x + x. a) (5.) Write the equation of the line tangent to the graph of f when x = 9. b) (8.3) Suppose you are using Newton s method to find a solution of the equation f(x) = 5. If your initial guess is x = 4, find x. c) (8.) Find the quadratic approximation to f when a = 4, and use this quadratic approximation to estimate f(5). 7. (.3) Solve the following initial value problem: { dy dx = 3x5 y 3 y() = 8. (7.4) Choose one of the following two problems: a) A wood box with a square base (the box has a top and a bottom) is to be made using 4 square units of plywood. What is the largest possible volume of such a box? b) A box is to be made using the following procedure: start with a piece of cardboard measuring inches by 4.5 inches. From each corner of the piece of cardboard, cut a small square of size x by x inches. Then take the remaining tabs and fold them up to form a box (with no top). If the box is to have the largest possible volume, what should the value of x be? 48

49 .. Fall 03 Final Exam Solutions. Usage of L Hôpital s Rule below is indicated with a L =: x 3 x + 4 (a) lim x x + 4 x + (b) lim x 5 = 3 () = 8 8 =. 5 x = 7 +0 =. x x 4 (c) lim x 4 x + 7x + = 0 L x = lim 0 x 4 x + 7 = 0 = 0. (This limit could also be done by factoring and canceling.) cos(4x) (d) lim = 0 x 0 x 0 (e) lim x 3 x 3 x = lim x 3 L 4 sin 4x = lim = 0 x 0 ( x 0 x 3) ( x ) L = lim x 0 6 cos 4x = 6 3x(x ) 3x(x ) = lim 3(x ) x(x ) x 3 3x 3x(x ) = lim (3 x)(x )3x( (x )) = lim (3 x)(x )3x(3 x) x 3 x 3 x = lim x 3 3x = 9. (f) lim 7e 4x = 7e = 7 x e = 7 = 0. = 8.. a) By the Product Rule, f (x) = 6x sin x + x 3 cos x. Now applying the Product Rule to each of these terms, we get f (x) = x sin x+6x cos x+ 6x cos x x 3 sin x = x sin x + x cos x x 3 sin x. b) First, rewrite the numerator using exponent rules to get y = 3x3/. Now, cot x by the Quotient Rule, y = 3 3 x/ cot x ( csc x)3x 3/ = 9 cot x cot x+3x 3/ csc x. x cot x c) Differentiate implicitly: d dx y x cos y = 5 d (y x cos y) = dx (5) ] dy [ dx cos y x sin y dy dx = 0 ( + x sin y) dy dx = cos y dy dx = d) By the Chain Rule, f (x) = x 4 +ln x (4x3 + x ). cos y + x sin y. e) Differentiate term-by-term to get f (x) = ex + x sec x. 49

50 .. Fall 03 Final Exam 3. (a) f) By the Quotient Rule, h (x) = [(x +)(x)] 3 x [ 3 x /3 ](x +) ( 3 (note that the x) Chain Rule was used to find the derivative of the numerator). Now h () = [( +)()] 3 [ 3 ()]( +) = 3 = 9. ( 4x + 3 ) ( dx = 4x + x / + 3 ) dx = 4 x x x 3 x3 + 4x / + 3 ln x + C. (b) (c) π/ π/4 (4 cos x + sin x) dx = [4 sin x cos x] π/ π/4 = [4 0] [ 4( /) / ] = e 4x dx = 5e 4x + C (by the Linear Replacement Principle) 3 [ (d) (x ) dx = ] 3 (x )3 = 3 6 (53 ) 6 = 4 6 = 6 3. (This integral applies the Linear Replacement Principle.) (e) Let u = x 4 3 so that du dx = 4x3 and du = 4x 3 dx. Therefore 6x 3 dx = 4 du so 6x 3 cos(x 4 3) dx = 4 cos u du = 4 sin u + C = 4 sin(x 4 3) + C. (f) Let u = x 3 + so that du dx = 3x and therefore 3 du = x dx. Now the limits need to be changed to u-values: when x = 0, u = = and when x =, u = 3 + = 9. So 0 x 9 x 3 + dx = 3 u du = 9 3 ln u = 3 ln 9 3 ln = 3 ln 9 0 = ln a) lim f(x) = (vertical asymptote at x = 4) x 4 b) lim f(x) = 3 (the horizontal asymptote to the left) x c) x lim f(x) = (graph increases without bound to the right) d) lim f(x) = 3 (left-hand limit) x e) lim f(x) DNE (left- and right-hand limits unequal) x f) f() = (the height of the dot at x = ) g) lim x 7 f(x) DNE (oscillating discontinuity a.k.a. too many wiggles) h) 4 f(x) dx = (the area under the curve from x = to x = 4 is a rectangle with height 6 and width 4 = ) 50

51 .. Fall 03 Final Exam i) x = 6, x = 9 are local minima (there are others as well) j) x = 9, x = are inflection points (technically x = 9 is as well even though it doesn t look like most inflection points) 5. a) f(3) which is clearly positive. b) Since the graph is decreasing near x =, f ( ) is negative. c) Since the graph has a cusp at x = 6, f (6) does not exist. d) Since the graph is concave up near x = 5, f ( 5) is positive. e) Since the graph is linear near x = 3, f (3) is zero. f) Since the graph lies above the x-axis between 3 and 4, 4 3 f(x) dx is positive, and since reversing the limits of an integral reverses its sign, 3 4 f(x) dx is negative. g) lim x f(x) does not exist because the left- and right-hand limits are unequal. 6. a) When x = 9, y = f(9) = =, so the line passes through (9, ). The slope of the line is f (9) = + = 7, so by the point-slope formula 9 6 the tangent line has equation y = + 7 (x 9). 6 b) To solve the equation x + x = 5, first rewrite so that one side of the equation is zero to get x + x 5 = 0. So we let g(x) = x + x 5 and apply Newton s method to g. If x 0 = 4, then x = x 0 g(x 0) g (x 0 ) = = 4 5 = = c) Notice that f(4) = = 6; f (4) = + = 5; f (x) = 4 4 f (4) =. Therefore the quadratic approximation is 4 4 3/ = 3 4 Q(x) = f(a) + f (a)(x a) + f (a)(x a) and f(5) Q(5) = = (x 4) (x 4) Separate the variables and integrate both sides: dy dx = 3x5 y 3 y 3 dy = 3x 5 dx y 3 dy = 3x 5 dx y = x6 + C. 4 x 3/ so 5

52 .. Fall 03 Final Exam Now plug in the initial condition and solve for C. We get 8 = + C so C = 5 8. Therefore the particular solution is y = + C so = x a) Let x be the length and width of the box and let h be the height. We are told that the surface area is 4, so this means x + 4xh = 4. This is the constraint; solving this for h yields h = 4 x = x = 6 x. 4x x x Now for the utility, which is the volume. V = lwh = x h = x ( 6 x x) = 6x x3. To maximize this, take the derivative and set it equal to zero: V (x) = 6 3 x V (x) = 0 when 6 = 3 x, i.e. 4 = x, i.e. x = ±. Throw out x = ; that leaves x =. (This is a maximum because V () = 3() = 6 < 0.) Last, the question asks for the largest volume which is V () = 6() ()3 = 4 = 8. b) The volume of the box is the length times the width times the height. The length of the box is x since two notches are cut out; similarly the width is 4.5 x. The height is x because the tabs are folded up to make the height of the box. So the volume is V (x) = x( x)(4.5 x) = (x x )( 9 x) = 54x 33x + 4x 3. Our goal is to maximize this function on [0,.5] (the.5 comes from half the length of the shortest side; since you cut two notches from each side, the biggest possible value of x is.5). To maximize this, take the derivative and set it equal to zero. V (x) = 54 66x + x = (7 33x + 6x ) = (7 6x)( x) So V (x) = 0 when x = and x = 7/6 (throw out x = 7/6 because it is bigger than.5). Finally, test the critical points and endpoints: CP: x = V () = = 5. EP: x = 0 V (0) = 0. EP: x =.5 V (.5) = (.5)( (.5))(4.5 (.5)) = (.5)(7.5)0 = 0. So the maximum volume is 5 when x =. 5

53 .. Spring 04 Exam. Spring 04 Exam. (.4) Sketch the graph of a function f which has all of the following properties: lim f(x) = 3 x + lim f(x) = x f() = 3 lim f(x) = x x lim f(x) = f has exactly one vertical asymptote lim f(x) does not exist x 7. (.4) Shown below is the graph of some unknown function f. Use the graph to estimate the following (if the quantity does not exist or is ±, say so): a) f( 6) b) lim x 6 f(x) c) f(7) d) x lim f(x) e) lim f(x) x + f) lim f(x) x g) lim f(x) x h) lim f(x) x 6 i) lim f(x) x

54 .. Spring 04 Exam 3. (3.) In each part of this problem, you are given a function which is not continuous at x = 4. Describe the type of discontinuity each function has at x = 4, and justify your answer (by computing appropriate limits). a) b) f(x) = f(x) = { 3x if x < 4 x 6 if x > 4 { x + 3 if x < 4 x 5 if x > 4 c) f(x) = x + x 4 4. (3.) Evaluate any five of the following six limits. (a) lim x + x 3 3 4x + x (b) lim x 8 x 8 x 4 (c) lim x x(x + 3) 5x (x ) (d) lim x x 5x + 6 x + 6x 6 (e) lim x x + x + x + x + 4 (f) lim x sin ( ) πx 3 54