Solutions to Math 41 First Exam October 18, 2012

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1 Solutions to Math 4 First Exam October 8, 202. (2 points) Find each of the following its, with justification. If the it does not exist, explain why. If there is an infinite it, then explain whether it is or. ( 2 (a) x x 2 ) x (4 points) ( 2 x x 2 ) 2 x x (x + )(x ) x 2 (x ) x (x + )(x ) x + x (x + )(x ) (x ) x x x + = 2 (x + )(x ) (b) x ln(2 + x) x 2 + x 2 + 2x + 2 (4 points) We have to compute x ln(2 + x) x 2 + x 2 + 2x + 2 We know that x 2 + ln(2 + x) = since x 0 + ln(x) =. So as x goes to 2 from the right, ln(x + 2) becomes a very large negative number. We also know that x 2 + x = 2 by direct substitution. Thus x ln(x + 2) becomes a very large positive number as x approaches 2 from the right. Since x 2 + x 2 + 2x + 2 = 2, and dividing a large positive number by a positive number close to 2 still gives us a large positive number, we have that x ln(2+x) becomes a very large positive number as well. Therefore, x 2 +2x+2 x 2 + x ln(2 + x) x 2 + 2x + 2 =

2 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 2 of (c) x 0 (sin 2 x) 2 cos(/x) (4 points) Since the it of cos(/x) as x goes to 0 doesn t exist, but cos(x) is always between - and, we will bound the function above and below and then use the Squeeze Theorem. cos(/x) 2 2 cos(/x) cos(/x) 2 2 sin2 (x) sin 2 (x)2 cos(/x) 2 sin 2 (x) Note that we can write the last inequality because sin 2 (x) is always greater than or equal to 0. From the last inequality, we see that x 0 2 sin2 (x) sin 2 (x)2 cos(/x) 2 sin 2 (x) x 0 x 0 We have that x 0 2 sin2 (x) x 0 2 sin 2 (x) = 0 by direct substitution. Therefore, by the Squeeze Theorem, x 0 sin2 (x)2 cos(/x) = 0 A common mistake was to try to find an inequality with sin 2 (x) in the middle. This doesn t work because the it of cos(/x) as x approaches 0 does not exist.

3 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page of 2. (0 points) Mark each statement below as true or false by circling either TRUE or FALSE. No justification is necessary. (a) For any positive ɛ and any x, we have TRUE FALSE (x 2 2x + 5) 4 < ɛ whenever 0 < x < ɛ. (2 points) If x < ɛ, then x 2 < ɛ. By algebra, the latter statement is equivalent to the statement (x 2 2x + 5) 4 < ɛ. (Remark: Think about how this statement could be used to prove that (x 2 2x + 5) = 4 using the delta-epsilon definition of its.) x (b) For any x, we have x 2 < whenever 0 < x 2 4. TRUE FALSE (2 points) If 0 < x 2 < 4, then x lies within /4 unit of /2 (though also x /2); that is, we know x lies in the interval (/4, /4). For such x, x 2 lies in (/6, 9/6), so x 2 lies between 5/6 and 7/6. Since this interval lies within (, ), we have x 2 <. (c) For any positive ɛ, there is a corresponding positive δ such that x 2 < ɛ whenever 0 < x 2 < δ. TRUE FALSE (2 points) Note that by our delta-epsilon definition of its, the above sentence is equivalent to the statement x 2 =, x 2 which is false! Thus, the original statement must be false. (To see this more directly, pick ɛ = 0.5. Assume such a δ exists. Pick x = 2 + γ where γ is less than δ and less than 0. but strictly larger than 0. We have that 0 < x 2 < δ, but x2 will be less than 0.6 = This ensures that x 2 > 0.64, which is a contradiction; thus there is no possible δ for this value of ɛ, and the original statement is indeed false.) Remark: Observe that part (b) can be true even though part (c) is false. (Part (b) is a special case of part (c), for one particular ɛ, that just happens to be true.) (d) For any function f and any a, if f(x) and f(x) x a + x a both exist and are equal, then f is continuous at x = a. (2 points) We also need f(a) to exist and be equal to f(x). x a (e) For any odd function f, if f(0) x 0 + f(x), TRUE TRUE FALSE FALSE then f is continuous at x = 0. (2 points) For such an f, since f is odd and defined at 0, we must have f(0) = 0. Thus, we have f(x) = 0. x 0 + Also, again using the oddness of f, we have f(x) f( x) f(x) = 0 = 0. x 0 x 0 + x 0 + Thus, we conclude f(x) = 0 = f(0), and so f is continuous at x = 0 by definition. x 0

4 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 4 of. (2 points) Let f(x) = ex + e x e x e x. (a) Find the equations of all vertical asymptotes of f, or explain why none exist. As justification for each asymptote x = a, calculate both the one-sided its f(x) and f(x), showing your x a + x a reasoning. (6 points) To find the vertical asymptotes of f(x) = ex +e x we must first find where it is e x e x discontinuous and in the case of the above quotient of continuous functions, it is therefore sufficient to determine where f is undefined; this will occur only when the denominator is 0, and: e x e x = 0 e x = e x x = x x = 0 Thus the denominator is 0 only if x = 0. To find out whether f(x) really has an asymptote at 0, we must take x 0 + ex +e x e x e x and x 0 ex +e x e x e x. As x approaches 0 from either side, the numerator, e x + e x approaches 2 and the denominator approaches 0. Thus the its are either or. To determine the signs, we must find out when e x e x is positive and negative. e x e x > 0 e x > e x x > x x > 0 Thus e x e x is positive when x > 0. Likewise, it is negative when x < 0. Thus, x = 0 is an asymptote, and e x + e x x 0 + e x = and e x e x + e x x 0 e x = e x (b) Find the equations of all horizontal asymptotes of f, or explain why none exist. Justify using it computations. (6 points) The function f(x) has a horizontal asymptote when at least one of its its as x approaches positive and negative infinity is finite. So we compute: x e x + e x e x e x x x = e (e x + e x ) x e (e x e x ) x + e 2x e 2x since x e x = 0. Note that we had to divide the top and bottom by e x because x e x =. Next we compute: e x + e x x e x e x = x x = (e x + e x ) e x (e e x e x ) x e 2x + e 2x since x e x = 0. Note that this time we had to divide top and bottom by e x since x e x =. Thus, the horizontal asymptotes are y = and y =.

5 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 5 of 4. (8 points) Prove, using precise statements, that there is a real number x between and which is a solution to the equation ( πx ) sin = ln(x 2 ). 2 The main difficulty of this problem is that ln(x 2 ) has a vertical asymptote at x = 0, so it is not continuous on [, ]. Arguments that did not take this into account received a maximum of 4 out of 8 points. Let f(x) = sin( πx 2 ) ln(x2 ). Since it is a combination of polynomials, trig functions, and log functions, it is continuous on its domain, which is (, 0) (0, ). We are looking for a value of x between and for which f(x) = 0. With the picture in mind, we focus on the interval [, 0). We have f( ) = sin( π/2) ln() = 0 =, and [ f(x) sin x 0 x 0 ( πx 2 ) ] ln(x 2 ) = 0 ( ) =. Since < 0 <, the Intermediate Value Theorem says that there must be some x in the interval (, 0) for which f(x) = 0, and therefore sin( πx 2 ) = ln(x2 ). Technically, we only learned the Intermediate Value Theorem for closed intervals. To be more rigorous, we d have to pick a small negative number b to be the right endpoint of our interval. One possible choice is b = /e. Then ( ) ( ) π π f(b) = sin ln(e 2 ) = sin ( 2) + 2 =. 2e 2e Since f( ) = < 0 and f(b) > 0, the Intermediate Value Theorem on the closed interval [, b] says that there must be some x in (, b) for which f(x) = 0.

6 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 6 of 5. (8 points) Let f(x) = 2 + x +. Find a formula for f (x) using the it definition of the derivative. Show the steps of your computation. We compute f f(x + h) f(x) (x) h 0 h 2+ (x+h)+ 2+ ( x+ (2 + (x + h) + )(2 + ) x + ) h 0 h (2 + (x + h) + )(2 + x + ) (2 + x + ) (2 + x + h + ) h 0 h(2 + x + h + )(2 + x + ) ( ) x + x + h + x + + x + h + h 0 h(2 + x + h + )(2 + x + ) x + + x + h + (x + ) (x + h + ) h 0 h(2 + x + h + )(2 + x + )( x + + x + h + ) h h 0 h(2 + x + h + )(2 + x + )( x + + x + h + ) h 0 (2 + x + h + )(2 + x + )( x + + x + h + ). We can plug h = 0 into this last expression, getting f (x) = (2 + x + )(2 + x + )( x + + x + ) = 2(2 + x + ) 2 x +.

7 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 7 of 6. (2 points) The reproduction pattern of a certain species of fruit fly, grown in bottles in a laboratory, depends on the number p of female flies in the bottle. A researcher determines values of S(p), the number of daily offspring per female; the chart below shows a few values of S(p): p S(p) (a) Give your best estimate for the value of S (4), showing your reasoning, and make sure to specify the units of this quantity. (4 points) By the definition of it, S f(q) f(4) (4) q 4 q 4. However, we are only given the values of S for discrete values of p, so for the best estimate we should average the values of the difference quotient for the two values of q nearest to 4: q = 2 and q = 6. S (4) = ( ) f(2) f(4) f(4) f(6) + = ( ) = The units of S are offspring per female fly per total number of female flies. (b) What is the practical meaning of the quantity S (4)? Give a brief but complete one- to twosentence explanation that is understandable to someone who is not familiar with calculus. ( points) S (4) describes the instantaneous rate of change of the number of daily offspring per female fly with respect to the number of female flies, when the number of female flies is 4. In practice this means that if we have a bottle with 4 female flies and we were to add a small quantity a of female flies, then the change in the production of daily offspring per female fly would be a. A lot of students answered this question by saying that if we were to add female fly to a bottle where we initially have 4 female flies, then the production of daily offspring per female fly would decrease by offspring per female fly. This not entirely correct, because female fly is not considered a small quantity when compared to the 4 female flies in the bottle. (One reasonable example of adding a small quantity : having 00 bottles with 4 female flies in each bottle, and then adding one female fly to exactly one of these bottles, which could be regarded as adding 0.0 female flies. In such a case, one could argue that we d expect to observe about S(4) S (4) daily offspring per female.)

8 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 8 of (Problem 6 continued) For easy reference, here again is the setup of the problem: The reproduction pattern of a certain species of fruit fly, grown in bottles in a laboratory, depends on the number p of female flies in the bottle. A researcher determines values of S(p), the number of daily offspring per female; the chart below shows a few values of S(p): p S(p) (c) For this and part (d), let g(p) = p S(p). Compute a formula for g (p). (Express your answer in terms of quantities such as p, S(p), and S (p).) (2 points) g(p) = p S(p). We are asked to compute the derivative of g with respect to p! Using the product rule we get: g (p) = p S(p) + p S (p) = S(p) + p S (p) NOTE: p is not a constant! g is a function of p only, so when we are asked for the derivative of g, we compute the derivative with respect to p. (d) State a practical implication of the statement g (4) is positive using only terminology of the laboratory setting. (You don t have to estimate the value of g (4); just assume it is positive for the purposes of answering this part.) ( points) The function g represents the number of offspring per female fly times the number of female flies which is the total number of offpring(per bottle per day). g (4) > 0 means that the function g is increasing near the point p = 4. The practical meaning of that is the following: if we start with a bottle with 4 femaly flies and we add a small quantity of female flies(see point b) for an explanation), the total number of offspring produced in this bottle will increase(even if the number of offspring per female decreases according to part b).

9 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 9 of 7. (8 points) Find the derivative, using any method you like. You do not need to simplify your answers. (a) h(x) = x + πx /4 (e/x) x (4 points) First, simplify h(x) as h(x) = x + πx 4 (e/x) x ( ) = x + πx 4 ex x 2 =x 2 + πx 4 2 ex 2 =x πx 4 ex 2. Use the power rule (x n ) = nx n to take derivative: h (x) = 5 2 x 2 π 4 x ex 5 2. sin x 2 cos x (b) f(x) = e x (4 points) Using the quotient rule, we have f (x) = (sin x 2 cos x) e x (sin x 2 cos x) (e x ) (e x ) 2 There is no need to simplify the solution further. = ((sin x) 2(cos x) ) e x (sin x 2 cos x) (e x ) e 2x = (cos x + 2 sin x) ex (sin x 2 cos x) (e x ) e 2x.

10 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 0 of 8. (7 points) The figure below shows the graph of a function f that has continuous first and second derivatives. The dashed lines are tangent to the graph of y = f(x) at (, ) and (5, ). List the following quantities in increasing order (from smallest to largest). No justification is necessary. f() f() f(2) (f(5) f(2)) The number 2 f ( 2 ) f () f (5) We first analyze the sign of the seven quantities. Since the graph has a horizontal tangent at x = 5, we know f (5) = 0. (f(5) f(2)) is the slope of secant line between x = 2 and x = 5, which is positive. Similarly, f() f(2) is the slope of secant line between x = 2 and x =, also positive. We also have f() =, positive. As for f (), f ( 2 ), since the slope of tangent lines at x = and x = 2 are negative, f (), f ( 2 ) are both negative. So far, we know f ( 2 ) f () 2 < f (5) = 0 < f() f() f(2) (f(5) f(2)) We next analyse the three negative terms. From the graph, the slope of tangent line at x = is 2 2 =. Although we do not know the exact slope of tangent line at x = 2, between x = 0 and x = the curve becomes steeper as we move forward alone the x-axis. This tells us f () is more negative than f ( 2 ): f () < f ( 2 ). We therefore have 2 < f () < f( 2 ) < f (5) < f() f() f(2) (f(5) f(2)) We now move on to the three positive terms. From the graph, f() f(2) is greater than 4, and (f(5) f(2)) is less than /. f() =. Therefore, we conclude 2 < f () < f ( ) < f (5) < (f(5) f(2)) < f() < f() f(2) 2

11 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page of 9. (2 points) The following is a graph of the function g: (a) Consider the function g, the derivative of g. Based on estimating g from the graph of g above, on what intervals is g increasing? decreasing? (No justification is necessary.) ( points) g is increasing if the graph of g is concave up, and g is decreasing if the graph of g is concave down. Referring to the graph, g is concave up on (, 0) and (2, ). Between x = and x = 2, g is concave down from x = 0 up until an inflection point x = a, and then g becomes concave up all the way toward x = 2. From the graph, one might estimate that a 4 ; at minimum one should find a value satisfying 2 a. Thus, the answer is g is increasing on (, 0), (a, 2), (2, ), g is decreasing on (0, a), where we gave full credit to any estimated a falling in the range 2 a. (b) Based on the picture of g above, which of the following expressions is a plausible formula for g(x)? Circle your answer; no justification is necessary. (You may take it as a given that exactly one of these formulas is the best answer.) x x 2 (x 2) x (x ) (x ) 2 x / (x 2) 2 x 2 (x 2) x / (x 2) 2 ( points) The four choices are of the form (x )m, where m, n are integers and b is either 2 or x b (x 2) n. We determine m, n, b by analysing the sign of g. When x is close to two, g is always positive. This says that moving from x > 2 to x < 2 does not change the sign of g(x). Therefore, n must be an even integer. Similarly, when moving from x > to x <, the function g(x) changes its sign from positive to negative, so m must been an odd integer. At this point there is only one candidate left: x x (x 2) 2 Let s check near x = 0 to see if this candidate really fits the given graph. Moving from x > 0 and x < 0 changes the sign of g(x). This is consistent with the choice b =.

12 Math 4, Autumn 202 Solutions to First Exam October 8, 202 Page 2 of (c) On the set of axes below, sketch a plausible graph of a function f satisfying all of the following: f is continuous on (, 2) (2, ), f (x) = g(x) for all x in the domain of g, f has vertical asymptote x = 2, and f(0) = 0. (6 points) Since g(x) = f (x) we have: (i) f is increasing when g > 0, decreasing when g < 0. (ii) f is concave up when g = f is increasing, concave down when g = f is decreasing. Summarizing, we have the following table. x 0 2 in/decrease concavity up up up down We now focus on the three points x = 0,, 2. f has to be continuous at x = 0 while f (x) goes to infinity as x approaches zero. Thus we know the graph of f has a cusp at x = 0. Put it another way, although the function is continuous at x = 0, the curve becomes steeper and steeper as we move toward x = 0 from both sides, and the tangent line finally becomes vertical at the point x = 0. Also, f(0) = 0. At the point x =, since f () = g() = 0, the curve has a horizontal tangent line. The line x = 2 is set to be a vertical asymptote. Hence we must have f(x) approaching infinity as x approaches 2. Remark: The x > 2 part of the graph does not have to be the same as the one drawn here. It can be translated vertically by any amount. The specification f(0) = 0 will not be affected by translating the red curve since the domain has already been broken into two parts: x < 2 and x > 2.