APPM 1345, Fall 2013: Eam 1 September 25, 2013 Instructions: Please show all of our work and make our methods and reasoning clear. Answers out of the blue with no supporting work will receive no credit. No calculators, notes or electronic devices are allowed. Please start each new problem on a new page in our bluebook. Please sign the front of our blue book indicating ou read and understood these directions in addition to the CU honor code. 1. (20 points) The graph of the derivative f of a twice-differentiable function f is shown here: Assume the relevant domain is [0, 8]. Answer the following with brief justification: (a) On what interval(s) is f increasing or decreasing? (b) At what value(s) of does f have a local maimum or minimum? (c) On what intervals is f concave upward or donward? (d) State the -coordinate(s) of an inflection points. (e) Assuming that f(0) = 0, sketch a graph of f, including an relevant point(s). # 1 (a) f is increasing on [1, 7] since f () > 0 on (1, 7) f is decreasing on [0, 1] and [7, 8] since f () < 0 on (0, 1) and (7, 8). (b) f has a local minimum at = 1 since f () changes sign from negative to positive there f has a local maimum at = 7 since f () changes sign from positive to negative there (c) f is concave up on (0, 2) and (4, 6) since f is increasing there (and so f () > 0) f is concave down on (2, 4) and (6, 8) since f is decreasing there (and so f () < 0) (d) f has inflection points at = 2, 4, 6 since f changes concavit at those -values (e) The graph should look something like this:
2 1 2 3 4 5 6 7 8 2. (33 points) In this problem ou will analze the curve = f() given the following function and its derivatives: f() = (3 + 1)2 ( 1) 2 f () = 8(3 + 1) ( 1) 3 f () = 48( + 1) ( 1) 4 Please provide the following information, being sure to organize our solution. That is, work for part (a) should be done when answering part (a), work for part (b) should be done when answering part (b) and so on. (a) What is the domain of f? (b) What are the intercepts? (c) Find the horizontal and vertical asmptotes, if an. You must justif our answers with the appropriate limits. (d) On what intervals is f increasing or decreasing? (e) At what -values does f have a local maimum or minimum? Find the corresponding -values, if an. (f) On what intervals is f concave upward or downward? Find an inflection points. (g) Finall, use our answers from parts (a) (g) to sketch a graph of the curve = f(), labeling all relevant information clearl. # 2 OVER (a) f is a rational function, so domain restrictions onl occur when the denominator is zero. Obviousl = 1 is the onl problem, so the domain is all real numbers ecept 1 (b) Since 0 is in the domain of f, the -intercept is f(0) = (3 0 + 1)2 (0 1) 2 = 12 ( 1) 2 = 1 So the point (0, 1) is on the graph.
3 To find the -intercept(s), we set the numerator equal to zero and solve: (3 + 1) 2 = 0 3 + 1 = 0 3 = 1 = 1 3. So the point ( 1/3, 0) is on the graph (c) Since f() = 92 + 2 +, lim f() = 9 ± 1 = 9 is the horizontal asmptote. (the degrees are the same on the top and bottom), so Note that the numerator (3 + 1) 2 4 2 = 16 as 1, whereas the denominator ( 1) 2 0 (through positive values) as 1 (from either side). It follows that lim f() = +, 1 ± so that f has a vertical asmptote at = 1. In other words, f() + as 1 ±. (d) f () is the product of three factors that could influence its sign: the 8, the (3 + 1) and the ( 1) 3. The last two factors are zero when = 1, 1, respectivel, so we test: 3 interval test point sign of f so f is (, 1/3) ( 1/3, 1) f ( 1) = 8(3( 1)+1) ( 1 1) 3 f (0) = 8(3(0)+1) (0 1) 3 (1, ) f (2) = 8(3(2)+1) (2 1) 3 It follows that f is decreasing on (, 1/3) and (1, ) whereas f is increasing on ( 1/3, 1). = 8( 2) ( 2) 3 < 0 (three negatives) DEC = 8(1) ( 1) 3 > 0 (two negatives) + INC = 8(7) 1 3 < 0 (one negative) DEC (e) From part (e), f has a local minimum at = 1/3 b the First Derivative Test. This -value is a critical point, and the corresponding -value is zero (see part (b)). The -value 1 is not in the domain of f and so no relative etremum occurs there (although increase/decrease changes there) (f) To analze concavit, note that f () and the factor ( + 1) in its numerator have the same sign. (The ( 1) 4 on the bottom is alwas positive!) Obviousl f () = 0 when = 1, and since it follows that + 1 < 0 when < 1 whereas + 1 > 0 when > 1, f () < 0 when < 1 whereas f () > 0 when > 1. Hence f is concave downward on (, 1) and f is concave upward on ( 1, ) (ecluding = 1)
4 Since concavit changes at = 1 (and this point is in the domain of f), f has an inflection point there. The corresponding -value is So the inflection point is ( 1, 1). f( 1) = (g) Finall, the graph should look something like this: (3( 1) + 1)2 ( 1 1) 2 = ( 2)2 ( 2) 2 = 1. 9 1 1 1 3 1
3. (27 points) A new smartphone is being designed that will be shaped like a rectangle cm long that is capped b semicircular regions of radius r at the top and bottom, as shown in the picture. If the outer perimeter of the phone will measure 12π cm, find the values of r and that will maimize the area of the viewing screen (the shaded rectangular region). r The outer perimeter of the phone is given b 2 + 2πr (the two edges plus one circumference). B the given (1) 2 + 2πr = 12π = + πr = 6π = = 6π πr. We want to maimize the area, A, of the shaded rectangle. But A = (2r). Substituting in for from (1) ields A(r) = 2r(6π πr) = 12πr 2πr 2. Note that r 0 is necessar, and b (1) r would be largest when = 0, which gives r = 6. So 0 r 6. If ou don t like this (because ou get a degenerate phone with no viewing screen, although the endpoints are mathematicall possible), one can also proceed assuming 0 < r < 6, but the solution is slightl different. Note that A(r) is differentiable and continuous everwhere since it s a polnomial in r. So the onl critical points occur when A (r) = 0. But A (r) = 12π 4πr, which is zero when Using the closed interval, 0 = 12π 4πr 4πr = 12π r = 12π 4π = 3. A(0) = 12π(0) 2π(0) 2 = 0, A(3) = 12π(3) 2π(3) 2 = 36π 18π = 18π, A(6) = 12π(6) 2π(6) 2 = 72π 72π = 0, which shows that an absolute maimum occurs at r = 3 b the Etreme Value Theorem/Closed Interval Method.
6 Alternativel, using the open interval, since A (r) = 4π(3 r) in factored form, it follows that A (r) > 0 if r < 3 whereas A (r) < 0 if r > 3, so that b the First Derivative Test A has a local maimum at r = 3. Since this is the onl critical point in the interval, we get an absolute maimum. In an case the area of the shaded rectangle is largest when r = 3. To find the corresponding, we use (1) again: = 6π π(3) = 3π. So the values of r and that ield the largest viewing screen are r = 3 and = 3π. 4. (20 points) This problem involves Newton s method. (a) We want to approimate the solution to the equation 3 = 3. Set up the general formula for calculating n+1 via Newton s method in this situation. (b) Calculate 2 given an initial seed of 1 = 1. Show our work and simplif our answer. (c) Now write down how ou would start calculating 3. You don t have to simplif this. (d) Finall, forget about parts (a) (c) and consider the following graph: 1 Cop this graph into our bluebook (ou could trace it). Show graphicall what 2 and 3 would be if ou applied Newton s method twice with 1 as a seed. What do ou think the root we re looking for is (don t think too hard)? Do ou think Newton s method is going to find it? Wh or wh not? # 4 (a) Rewrite the equation as 3 + + 3 = 0. Let f() = 3 + + 3.
7 Since f () = 3 2 + 1, the general formula for n+1 would be n+1 = n f( n) f ( n ) = n 3 n + n + 3 3 2 n + 1 (b) With 1 = 1, 2 = 1 13 + 1 + 3 3(1) 2 + 1 = 1 5 4 = 4 4 5 4 = 1 4 (c) With 2 = 1, we would start finding 4 3 like this: 3 = 1 ( ) 1 3 ( 4 4 + 1 4) + 3 3 ( 4) 1 2 + 1 and we can stop there. (d) The graph should look something like this: 2 1 3 The root looks like = 0 from the original graph, but Newton s method doesn t appear to to be finding it since 2 is farther from = 0 than 1 is, and 3 is father awa from = 0 still. It looks like the n s will just keep getting farther awa from = 0 as n increases since the graph flattens out farther to the left and right. Some people thought that 3 landed right on top of 1, so that the n s would just keep bouncing back and forth between two fied values. That was acceptable since our graph (even if ou traced it) might have looked that wa. The point was that Newton s method fails.