Solutions to Math 41 First Eam October 12, 2010 1. 13 points) Find each of the following its, with justification. If the it does not eist, eplain why. If there is an infinite it, then eplain whether it is or. 5 e ) a) 5 2 6 + 5 4 points) We manipulate the it: 5 e ) ) e 5 2 6 + 5 5 5 2 by rearranging) 6 + 5 ) e 5 5 by factoring) 1) 5) ) e 5 1 by cancellation, because 5). 1) We may now substitute in = 5, which yields the it: e 5 5 1 5 1) = e5 20. b) 1 + 4 2 3 5 points) We manipulate the it: 1 + 4 2 3 1 2 3 + 1 2 1 1 3 2 4 ) 2 3 2 + ) + 1 2 4 4 2 3 4 simplifying). by splitting up terms) ) multiply each term by 1) Note that whereas Hence, our desired it is 1 1 1 2 1 1 =, ) + 1 2 1 2 1 1 ) = 0. =.
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 2 of 12 c) cos sin 1 ) 0 + 4 points) Let f) = cos sin 1 ) ; we may then rewrite cos sin 1 ) f). 0 + 0 + Since we have 1 cosθ) 1 for all θ, 1 f) 1 for all. As 0 +, we may assume is positive and therefore that eists; hence, we multiply all three sides to obtain f). Since the two its 0 + and 0 + both eist and equal 0, the squeeze theorem says that also eists and is 0. f) 0 +
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 3 of 12 2. 15 points) Let f) = 4e + 1 2e 3. a) Find the equations of all vertical asymptotes of f, or eplain why none eist. As justification for each asymptote = a, compute at least one of the one-sided its showing your reasoning. f) or a + a f), 5 points) Being built from continuous functions, f) is continuous on its domain, which includes all real numbers for which the denominator is non-zero. So the domain is, ln 3 2 ) ln 3 2, ), and the only possible vertical asymptote is = ln 3 2. Let s check by calculating one of the onesided its, say 4e + 1 f) ln 3 2 ) ln 3 2e 3 2 ) Since the numerator is greater than 1, and the denominator is an increasing function of crossing zero at = ln 3 2, the it involves infinity. Furthermore, we see that f changes sign from negative to positive at = ln 3 2, so the it from the left is. Similarly, the it from the right is + ). In conclusion, f has eactly one vertical asymptote, = ln 3 2. b) Find the equations of all horizontal asymptotes of f, or eplain why none eist. Justify using it computations. 5 points) To find the horizontal asymptotes we investigate the its of f) as tends to ± : f) 4e + 1 2e 3 = 4 0 + 1 2 0 3 using it laws and e = 0) = 1 3 f) 4e + 1 + + 2e 3 4 + e + 2 3e = 4 + 0 2 3 0 = 2 using it laws and + e = 0) Both its eist and are finite, so f has two horizontal asymptotes, y = 1 3 and y = 2.
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 4 of 12 For easy reference, recall that f) = 4e + 1 2e 3. c) It is a fact that f is a one-to-one function. Find an epression for f 1 ), the inverse of f. 5 points) We find the epression using a series of algebraic manipulations, starting with y = f 1 ), remembering that y ln 3 2 : y = f 1 ) = fy) = 4ey + 1 2e y 3 2e y 3) = 4e y + 1 2e y 3 = 4e y + 1 2 4)e y = 3 + 1 e y = 3 + 1 2 4 ) 3 + 1 y = ln 2 4 So the sought-after epression for the inverse is f 1 ) = ln ) 3 + 1. 2 4 assuming 2) assuming 3+1 2 4 > 0) In parentheses, we remark that the above calculation indeed shows that f is one-to-one with range, 1 3 ) 2, + ), this being the range of validity of the assumptions made on.)
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 5 of 12 3. 9 points) Find values of a and b so that the function a h) = 1 + 3 if < 2 b 2 4 if 2 is continuous on, ). Justify your answer. First of all, notice that if a 0, the function is not defined at = 1 and therefore, it is not continuous there. So we need to set a = 0 to make h) continuous at = 1. So now the function becomes: h) = { 3 if < 2 b 2 4 if 2 If < 2, h is a constant function near, so it is continuous at. If > 2, h is the difference of a polynomial and an eponential function near, which are continuous, and differences of continuous functions are continuous, so h is continuous at. Thus h is continuous at all 2, no matter the value of b. Let us now study continuity at = 2; that is, we need to find b such that h2) 2 h). Since h) is a piecewise function, we need to compute the one-sided its. f2) = 4b 16 h) 3 = 3 2 2 h) 2 + 2 b2 4 = 4b 16 + In order for these three values to be equal, we need to have 4b 16 = 3, that is, b = 19 4. Hence the only values of a and b that make h) continuous on, ) are a = 0 and b = 19 4. Notes: In discussing the continuity of h at points other than = 1, 2 which is essential for full credit!), many described h simply as a polynomial at all such points. But in fact the pieces of h should be treated separately and note 4 is not a polynomial function, though we have seen that it is continuous as an eponential function). Also note that we can t easily argue the continuity at = 2 in this same way, i.e., by saying that b 2 4 the formula for h on the piece that includes 2) is a continuous function for all. This would only be valid if h) is equal to b 2 4 for all points near = 2. But since h has a different formula at all points just to the left of = 2, we must instead appeal to the it definition of continuity at = 2 as eplained above).
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 6 of 12 4. 8 points) Let f) = 5 4. Find a formula for f ) using the it definition of the derivative. Show the steps of your computation. f f + h) f) ) it definition of derivative) h 0 h 1 + h 5 h 0 h + h 4 5 ) plug in f) 4 ) 1 + h 5) 4) + h 4) 5) common denominators) h 0 h + h 4) 4) ) 1 4) 5) + h 4) 4) 5) h 5) h 0 h + h 4) 4) ) 1 h 4 + 5) h 0 h + h 4) 4) ) 1 h h 0 h + h 4) 4) 1 cancel h from top and bottom) h 0 + h 4) 4) 1 = 4) 2 take it by direct substitution h = 0)
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 7 of 12 5. 6 points) Eplain completely whether g) = sin ) is differentiable at = 0, and if it is, find g 0). Your reasoning should be complete; in particular, a picture alone is not sufficient. { if 0 First, since =, we have g) = if < 0 { sin if 0, so g) = sin ) if < 0 { sin if 0 sin if < 0 where we use the fact that sin is odd. It turns out that g is not differentiable at 0; however, due to the piecewise nature of g, there is essentially only one way to eplain this. We must show that g 0) does not eist, by the it definition of the derivative. If g 0) did eist, then g g) g0) 0) 0 0 sin sin 0 0 0 sin 0 0 0 0 sin, but we ll see that this last it does not eist. Indeed, using the piecewise view of g, we have sin sin 0 + 0 + = 1 see below for reasoning), and ) sin = 1. 0 sin sin ) sin = 0 0 0 sin Thus, because the one-sided its are not equal to each other, the it does not eist. It 0 follows that g 0) does not eist; equivalently, g is not differentiable at 0. sin For completeness, there are a couple of ways to eplain why = 1. For one, this is a fact 0 proved in the tetbook reading via the Squeeze Theorem and right-triangle trigonometry); you could mention this and receive full credit. We also gave full credit to the following use of the it definition of the derivative: sin sin 0 sin 1 = cos 0 = [derivative of sin, evaluated at 0] 0 0 0 sin h Typically, though, the derivative of sin is computed using the fact = 1, not vice versa.) h 0 h Notes: Many people confused the notions differentiable at 0 and continuous at 0. In fact g is continuous at = 0, being a composition of functions continuous at all. But differentiability is a more restrictive condition than continuity, and it is determined through an entirely different process i.e., by eamining the above difference quotient it, not by asking if g) = g0)). 0 Other responses cited the non-differentiability of at = 0 as an automatic reason for the nondifferentiability of g, a composition involving, at 0. But there is no such automatic reasoning possible, since cos ) and ) 3 are both functions that are differentiable at = 0 check this!). { Other responses wrote that g ) = cos, but this is false. In fact, g cos if > 0 ) = cos if < 0. Other responses, after correctly computing g ) at nonzero, noted that g ) does not eist 0 and this is true); but unfortunately, this fact alone does not have bearing on whether g 0) eists. Note that the differentiability of g at 0 depends only on the eistence of g 0).) See #3 on the Fall 2008 final eam for an eample of a function f, differentiable at 0, for which f ) does not eist. 0 Finally, here is a graph of g it was not required to give a graph, and in fact many gave an incorrect one). The non-differentiability at = 0, proved above, is informally seen as a corner in this graph. 1-3π -2π -π 0 π 2π 3π -1
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 8 of 12 6. 10 points) Data taken from a certain speed camera outside Santa Fe, NM, shows that traffic speed S, in mph, varies as a function of the traffic density q in cars per mile) at that point on the road. The following table gives S for a few values of q: q 100 110 120 130 140 Sq) 46 43 40 38 37 a) Give your best estimate for the value of S 120), showing your reasoning, and make sure to specify the units of this quantity. 4 points) We have S Sq) S120) 120) q 120 q 120 Since only discrete values of the function Sq) are given, our best estimate of S 120) is found by averaging the values of this difference quotient for the nearest values of q to 120, namely for q = 110 and q = 130. S 120) 1 [ ] S110) S120) S130) S120) + 2 110 120 130 120 = 1 ) 43 40 38 40 + 2 10 10 = 1 3 2 10 2 ) 10 = 1 mph 4 cars per mile b) What is the practical meaning of the quantity S 120)? Give a brief but complete one- or twosentence eplanation that is understandable to someone who is not familiar with calculus. 3 points) S 120) describes the instantaneous rate of change of traffic speed with respect to traffic density, when the traffic density is 120 cars per mile. More concretely: when the traffic density is initially 120 cars per mile, S 120) gives us the ratio of a small change in traffic speed to the corresponding small change in traffic density. This allows us to estimate the predicted traffic speed for values of traffic density near 120 cars per mile.) c) The product V q) = q Sq) is called traffic volume. Find a formula for V q). Epress your answer in terms of other quantities, such as q, Sq), and S q).) 3 points) Since V q) = qsq) we can use the product rule to find V q): V q) = q Sq) + qs q) = Sq) + qs q) here we used the fact that the derivative of the function fq) = q is the constant function 1.
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 9 of 12 7. 6 points) The graph of the function g is given below, as well as the graphs of the function s first and second derivatives, g and g, respectively. Indicate which graph belongs to which function, and give your reasoning. Since we know the three curves are in fact the graphs of g, g, g in some order, we can use process of eination in our reasoning. First we look at the functions b and c. We see that b cannot be c, because b is increasing on the far right, whereas c is negative there. Similarly c cannot be b, because c has a local maimum near 0 whereas b is negative there. Thus, we are faced with either Case 1: b = g, c = g or Case 2: b = g, c = g. In Case 1, we would have b = a. But b is decreasing on the left side of the graph, whereas a is positive there; hence b a and Case 1 is impossible. Finally, by process of eination, we are left with the choice b = g, c = g, and hence a = g. Notes: There are other ways to justify this answer, but the answer itself is unique.
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 10 of 12 8. 12 points) Find the derivative, using any method you like. You do not need to simplify your answers. a) Rt) = t 7.3 + t4 + t t 4 points) Begin by noting that Rt) = t 7.3 + t 3 + t 1/2. Then R t) = 7.3)t 8.3 + 3t 2 1 2 t 3/2 b) f) = e e e + 3π 2 7 4 points) f ) = d d e ) e d d e ) + 3π 2 d 7 ) d = 1 e + e ) e 2 e 1 + 21π 2 6 c) P y) = cos y 1 y sin y 4 points) P y) = = ) ) y sin y) d dy cos y 1) cos y 1) d dy y sin y) y sin y) 2 y sin y) sin y) cos y 1)sin y + y cos y) y 2 sin 2 y Alternatively, rewrite P y) = y 1 )cot y csc y) and use the product rule: P y) = y 2 )cot y csc y) + y 1 ) csc 2 y + csc y cot y)
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 11 of 12 9. 15 points) The following is a graph of the function g: a) On the aes below, sketch a plausible graph of g, the derivative of the function graphed above. Label the scales on your aes. 5 points for part a), 2 points each for parts b) and c), 6 points for part d))
Math 41, Autumn 2010 Solutions to First Eam October 12, 2010 Page 12 of 12 For easy reference, here is the same graph again: New information: Now suppose instead that we think of the above as the graph of the derivative of some function h. Give brief answers to the following questions about the function h. You do not need to provide justification for your answers. But be careful: the questions below pertain to h, even though we are now supposing the above graph is a picture of h, the derivative.) b) On what intervals is h increasing? decreasing? h is increasing if and only if h is positive, so h is increasing on 6, 4). h is decreasing if and only if h is negative, so h is decreasing on, 6) and 4, 5) and 5, ). For the latter, we also accepted 4, ), since in part d) we take that h5) is defined.) c) On what intervals is h concave up? concave down? h is concave up if and only if h is increasing, which occurs on, 4) and 1, 3) and 5, 6). h is concave down if and only if h is decreasing, which occurs on 4, 1) and 3, 5) and 6, ). d) Now suppose additionally that h0) = 0 and that h is continuous on, ). Sketch a plausible graph of h on the set of aes below. Label your aes scales, and clearly label the locations of horizontal tangents HT ) and inflection points IP ). 6 points) The y-scale is very approimate; we know the slope of h at = 1 should be about 1, and the slope at = 3 should be more than 3, etc. HT at = 6, 4 IP at = 4, 1, 3, 5, 6 Note also that the graph must have a vertical tangent line at = 5. The precise location of the positive -intercept of h is difficult to know solely from the graph of h.)