FINAL EXAM 1 SOLUTIONS Below is the graph of a function f(x). From the graph, read off the value (if any) of the following limits: x 1 +

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1 FINAL EXAM 1 SOLUTIONS Below is the graph of a function f(x). From the graph, rea off the value (if any) of the following its: x 1 = 0 f(x) x 1 + = 1 f(x) x 0 = x 0 + = 0 x 1 = 1 1

2 2 FINAL EXAM 1 SOLUTIONS Big Bir ries the roller coaster at the Santa Cruz Boarwalk. After he boars the train, the train is lifte up the first big slope by a chain uner the track. The chain carries the train up at a constant spee. When the train reaches the top, it is release to go own the hill uner the force of gravity. Assume that the hill has a long straight slope, only rouning off very near the bottom to a flat section of track. We will only consier the up-an-own part of the rie, not the part after coming own the straight slope. Also we will only consier his height an his vertical velocity an acceleration. Assume that the boaring platform an the bottom en of the straight track are at the same height, an that the train is very short, so we can ignore the short time when the front of it is over the top but the back is still coming up. Solution. We have to consier the time before he reaches the top (say at t 0 ) an the time when he is coming own (after t 0 ). Since the chain carries the train up at a constant spee, his acceleration is zero for t t 0. Afterwars, his acceleration is a negative constant (somewhere between g an 0 epening on how steep the roller coaster is an how much or little friction there is, but all the matters to us is that it is a negative constant). Now that we know the acceleration, the velocity must be a positive constant for t t 0 an a ownwar-sloping line for t t 0. Then the position is an increasing line, followe by a concave-own parabola. (Ignore the part for t < 0 at the left of the following graphs. The top graph is height, followe by velocity an acceleration.)

3 3. Evaluate the following its. FINAL EXAM 1 SOLUTIONS x 2 (x4 + 3x + 1) Polynomials are continuous, so the it is = = 29. x 2 4 x 2 x 2 (x 2)(x + 2) = x 2 x 2 = (x + 2) x 2 = 4 x x sin x = x 0 x x 0 sin x sin x = x 0 x x 0 x = 1 x x 0 = 0 x x since the first it is 1 an x/ x = x ( ) t 3 t t ( ) t 3 = t t 3 1 t3 1 t 3 1 ( ) 1 = t t 3 1 = 0 since the it of a rational function at is etermine by the leaing terms

4 4 FINAL EXAM 1 SOLUTIONS (a) Fill in the blanks to make the following correct: To prove (from the efinition of it) that 3 x approaches 0 as x approaches zero: Let ɛ > 0 be given. Then choose δ = ɛ 3. Then we verify that if x < δ then 3 x < ɛ. (b) Draw a picture showing part of the graph of y = 3 x an labeling ɛ an δ. Here s the graph, but I on t have a way to raw ɛ an δ on this igital picture. It woul just be a rectangle with center at the origin an two vertices on the curve, an the with is 2δ an the height is 2ɛ. 5. For each of the following functions, if the function is not continuous at every x, at which x is it not continuous? If it is continuous everywhere then just write continuous everywhere. x 1 x 2 1 Discontinuous at x = ±1 because the enominator is zero, so the function is unefine. It oesn t matter that x 1 cancels. Continuous everywhere 3 x tan x Discontinuous where unefine, namely when x is an o multiple of π/2.

5 FINAL EXAM 1 SOLUTIONS Continuous everywhere e sin x x cos 1 x 6. Evaluate the following its. Solution : Solution : (4 x x) x (4 x x) = x x (x1/4 x 1/2 ) ( ) t 3 t t 1 t2 t = x (x1/4 (1 x 1/4 )) = x x1/4 (1 x x1/4 ) = ( ) = ( ) t 3 t t 1 t2 t 2 t 1 = t t 3 t 2 (t 1) t 1 = The last it can be one in one step because it is a it of a rational function whose numerator has larger egree than its enominator.

6 6 FINAL EXAM 1 SOLUTIONS Below is the graph of a function. Sketch the graph of its erivative on the secon, blank graph given. Every stuent knows how to sketch this graph, I think, so I i not spen the time require to make a igital image. 8. Evaluate the following erivatives x (x49 + 2x + 1) = 49x x ((x32 + 3x 3 )(sin x + 1)) = (x x 3 ) (sin x + 1) + (sin x + 1) x x (x32 + 3x 3 ) = (x x 3 ) cos x + (sin x + 1)(32x x 2 ) x x e x = ex x x x x ex (e x ) 2 = ex /(2 x) xe x (e x ) 2 = 1/(2 x) x) e x = 1 2x 2e x x

7 FINAL EXAM 1 SOLUTIONS x x = 3x2 (x 3 + 1) 2 9. Evaluate the following erivatives x (x3 sin x) = x 3 cos x + 3x 2 sin x 1 ln(ln x) = x ln x x ln x = 1 x ln x x sin(x8 ) = cos(x 8 ) x x8 = 8x 7 sin(x 8 ) (sin 2x cos x) x = sin 2x cos x + cos x sin 2x x x = (sin 2x)( sin x) + cos x cos 2x x 2x = sin x sin 2x + 2 cos x cos 2x 10. Suppose x 3 + tan y = 1. Calculate y/x using implicit ifferentiation, getting an answer that involves both x an y. Solution : x 3 + tan y = 1 3x 2 + sec 2 y y x = 0 y x = 3x2 sec 2 y = 3x 2 cos 2 y

8 8 FINAL EXAM 1 SOLUTIONS Eve of Naharon is a skeleton of a young female foun in an unerwater cave in Mexico. Raiocarbon ating one in 2008 showe that only 19.3% of Eve s original carbon 14 remains. What was Eve s eath ate, accurate to the nearest ten years? Here are some useful numbers: Half-life of carbon 14 is 5730 years; ln(2) = 0.693; log 2 (0.193) = ; ln(0.193) = The only acceptable answer is an integer for the year of the eath. Solution : (1/2) t/5730 = That is, 2 t/5730 = Taking logarithms to the base 2 we have t 5730 = log 2 (0.193) = t = So the skeleton is about 2.4 times 6000 years ol, or aroun 14 thousan years ol; since we nee it to ten years accuracy that is somewhat more accurate than one part in a thousan. So we nee to multiply 5730 by We get 13,597. Subtracting 2008 we fin the answer: 11,589 BCE.

9 FINAL EXAM 1 SOLUTIONS Two commercial airplanes are flying at 40,000 feet along straight-line courses that intersect at right angles. Plane A is approaching the intersection point at a spee of 442 miles per hour. Plane B is approaching the intersection at 481 miles per hour. At what rate is the istance between the planes changing when A is 5 miles from the intersection point an B is 12 miles from the intersection point? (a) Draw a iagram of this situation. Define three variables (other than t for time) an inicate their meaning on the iagram. y A z B x x is the istance of B from the intersection point, y is the istance of A from the intersection point, an z is the istance between the two planes. (b) What, in symbols, is the unknown quantity? z/t at a certain instant. (c) Write an equation connecting the variables that is goo all the time the planes are moving as escribe. x 2 + y 2 = z 2 () Finish solving the problem. 2x x y z + 2y = 2z t t t We are given x/t = 481 an y/t = 442. Both are negative since the planes are flying towars the intersection point. 481x 442y = z z t At the moment of interest x = 12 an y = 5; so from z 2 = x 2 +y 2 we have z = (144+25) = 169 = 13. (Hopefully you can figure that out without a calculator!) Putting these numbers in we have = 13 z t Doing the arithmetic we fin z t = = 614 The istance between the planes is ecreasing at a rate of 614 miles per hour, or changing at a rate of -614 miles per hour.

10 10 FINAL EXAM 1 SOLUTIONS At the Monterey Bay Aquarium, there is a system of large pipes to supply sea water to the big tanks. The pipes ten to become partly clogge with plant material, which restricts the flow of water, so from time to time they clean the pipes (exactly how is interesting but not relevant to this problem). A similar problem arises in oil pipelines, which also nee to be cleane out from time to time. The equation relating flow an raius is V = kr 4, where r the raius of the pipe, k is a constant, an V is the volume of water that can flow in one secon past a given point. Suppose that ue to stuff accumulating on the insie of the pipe, the raius ecreases by 4%. By what percent will the flow ecrease? Solution : V = kr 4 V = 4kr 3 r V = 4kr 3 r V = 4kr3 r V V 4kr = 3 r kr 4 since V = kr 4 = 4 r r We re given r/r = Therefore V/V is four times that, or 0.16, which is 16%. 14. Fin the numerical value of cosh(ln 10). Show your work, an be sure that your answer is a number. cos(ln 10) = 1 2 (eln 10 + e ln 10 ) = 1 2 (eln /e ln 10 ) = 1 2 (10 + 1/10) = 5.05

11 FINAL EXAM 1 SOLUTIONS Consier the function f(x) = x 3 4x 2 + 5x + 1. You can show your work below in the space provie for it, but please put your answers right after the questions where they will be easy to fin. (The question will be grae as a whole; o not assume that all the parts have equal value.) (a) For which x is f(x) increasing? (, 1) an (5/3, ) (b) For which x is f(x) ecreasing? (1, 5/3) (c) Fin the values of x an f(x) at the local minima of f. (5/3, 77/27) () Fin the values of x an f(x) at the local maxima of f. (1, 3) (e) Fin the points of inflection of f on the given interval. 4/3 is the only inflection point. (f) Where is f concave up? (4/3, ) (g) Concave own? (, 4/3) f (x) = 3x 2 8x + 5 f (x) = 6x 8 So f is zero when x = 4/3, an positive for x > 4/3, an negative for x < 4/3, so 4/3 is a (an the only) point of inflection, an we can answer (e) through (g). Setting f (x) = 0 we have 3x 2 8x + 5 = 0. Solving for x we fin x = 4/3 ± 64 60/6 = 4/3 ± 1/3 so the critical points are at x = 1 an x = 5/3. Since f is positive at 5/3 an negative at 1, there is a minimum at 5/3 an a maximum at 1. Plugging those values into the formula for f we have f(1) = = 3 f(5/3) = (5/3) 3 4(5/3) 2 + 5(5/3) + 1 = ( )/27 = 77/27 = Since f is a quaratic with positive leaing term, f is positive on (, 1) an (5/3, ), an negative on (1, 5/3). So we can answer parts (a) an (b) now.

12 12 FINAL EXAM 1 SOLUTIONS 2011 (h) Sketch the graph of f showing all the minima, maxima, an points of inflection.

13 16. Evaluate the following its: FINAL EXAM 1 SOLUTIONS x 0 x 3 ln x Since the numerator goes to zero an the enominator to, the answer is zero. We can t use L Hospital s rule here because this is not an ineterminate form, even though in this case L Hospital s rule woul accientally give us the right answer. ln(cos x) x 0 sin x Since both numerator an enominator go to zero, we can use L Hospital s rule: ln(cos x) x 0 sin x = x ln cos x x sinx sin(x)/ cos x = x 0 cosx = x 0 = 0 sin x cos 2 x ln(x ) = 0 x 0 ln x Since the numerator oes not go to ± it is not an ineterminate form. The numerator is continuous at 0 an its value there is 10; since this is not zero an the enominator goes to, the answer is 0. You cannot use L Hospital s rule on this problem.