Solutions to Math 41 First Exam October 15, 2013

Solutions to Math 41 First Exam October 15, 2013 1. (16 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. (a) x 2 ln(x/2) x 1 + x 2 1 (4 points) By substituting x = 1 we get ln(1/2) in the numerator, which is nonzero, and 0 in the denominator. This means the it involves infinity. So it is enough to find its sign. Observe that when x 1 + we get x 2 ln(x/2) ln(1/2) < 0 and x 2 1 > 0. So x 2 ln(x/2) x 1 + x 2 = 1 (b) z 2 z 2 4 z 2 + z 2 (4 points) Direct substitution gives the indeterminate form 0 0. So we need some algebraic manipulation: z 2 z 2 4 z 2 + z 2 (z + 2)(z 2) z 2 (z + 2)(z 1) z 2 z 2 z 1 = 2 2 2 1 = 4 3

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 2 of 11 (c) e x + 2 x x 3 x + 1 (4 points) Since x ax = 0 for all 0 < a < 1 we get 2 x + e x x 3 x + 1 x x (( 2 x ( ) e x 3 3) x + 3) ) x ) 3 x (1 + ( 1 3 ( ) 2 x ( e x + 3 3) 1 + = 0 + 0 1 + 0 = 0 ( ) 1 x 3 (d) t sin(ln t) t 0 + (4 points) Since 1 sin θ 1 for all θ, we have So 1 sin(ln(t)) 1 for all t > 0. t t sin(ln(t)) t for all t > 0. Since t = 0 t by the Squeeze Theorem we deduce that t 0 + t 0 + t sin(ln(t)) = 0 t 0 +

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 3 of 11 2. (8 points) (a) Complete the following sentence: the rigorous ( ɛ, δ ) definition of the statement f(x) = L x a is that for every positive value of, there is a corresponding positive value of so that whenever, then. (3 points)... is that for every positive value of ɛ, there is a corresponding positive value of δ so that Remark. Some common mistakes: whenever 0 < x a < δ, then f(x) L < ɛ. The expression 0 < x a < δ was often written simply as x a < δ, which is incorrect because the it only uses information near x = a but not actually at x = a. The implication was often written backwards: if f(x) L < ɛ, then 0 < x a < δ. That this is incorrect can be seen by considering the function f(x) = 0, which, of course, has the it L = 0 everywhere. Then using the backwards definition of the it, f(x) L = 0 0 = 0 < ɛ is always true and so imposes no constraint on x a. In particular, we can take x a as large as we like, so in fact no acceptable choice of δ < exists. (b) Use the ɛ, δ definition to prove that x 0 x 2/3 = 0. (5 points) Let ɛ > 0 be given. We want to find a value of δ > 0 so that whenever 0 < x 0 < δ, then x 2/3 0 < ɛ. But x 2/3 0 = x 2/3 = x 2/3, so the condition that x 2/3 0 < ɛ is equivalent to the condition that x 2/3 < ɛ, which itself is equivalent to the condition that x < ɛ 3/2. Thus, the choice δ = ɛ 3/2 works because if 0 < x 0 = x < δ = ɛ 3/2, then x 2/3 0 = x 2/3 < δ 2/3 = ɛ. Remark. Some common mistakes: Many proofs did not specify that ɛ is an arbitrary positive number. Without first defining what ɛ is, it makes no sense to talk about setting δ = ɛ 3/2. This is why the solution above begins with Let ɛ > 0 be given. This makes it clear that ɛ is some arbitrary number, which is necessary to satisfy the for every positive value of ɛ part of the it definition. Along those same lines, many proofs did not adequately explain what was going on. Proofs typically contain complete sentences, as above. It is generally not sufficient to just show calculations without explaining what you are doing. Some proofs used δ = ɛ x 1/3, obtained by multiplying the inequality x 2/3 < ɛ through by x 1/3. But this δ is not a number since x is a variable! However, one can combine the equation for δ with the inequality 0 < x < δ to write δ < ɛδ 1/3, from which it immediately follows that δ 2/3 < ɛ or δ < ɛ 3/2 exactly as above.

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 4 of 11 3. (14 points) Let h(x) = e ( ex ) 2. (a) Find the domain and range of h. (3 points) The domain of h is all of R = (, ), because h is a composition of exponential and linear functions, all of whose domains are R. To find the range of h: range of e x is (0, ), range of e x is (, 0), range of e ( ex) is (0, 1), range of e ( ex) 2 is ( 2, 1). OR From part d, h 1 (x) = ln( ln(x + 2)), and the range of h is the domain of h 1. For h 1 (x) to be defined, the arguments of both lns must be positive, i.e. x + 2 > 0 x > 2 and ln(x + 2) > 0 ln(x + 2) < 0 x + 2 < 1 x < 1. So the range is {x R 2 < x < 1}. OR Many students said that the range is ( 2, 1) because the horizontal asymptotes (found in part c) are y = 2 and y = 1. This received no credit since this approach usually finds the wrong answer: for example, f(x) = ex +x e x x has horizontal asymptotes y = 1 and y = 1, but for all positive x, f(x) > 1, so its range is not just ( 1, 1). If you really want to mend this argument with horizontal asymptotes, you can say: h (x) = d dx ( ex )e ( ex) = e x e ( ex) < 0 for all x, so h(x) is a decreasing function that is continuous on all of R, so its range is the interval between x h(x) and x h(x), i.e. the interval ( 2, 1). (b) Find the equations of all vertical asymptotes of h, or explain completely why none exist. (As justification for each asymptote x = a, calculate both the one-sided its h(x) and h(x), x a x a + with reasoning.) (2 points) h has no vertical asymptotes because it is continuous on all of R. OR The range of h is ( 2, 1), so h(x) cannot be arbitrarily large. Hence x a h(x) and x a + h(x) cannot be or.

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 5 of 11 For easy reference, h(x) = e ( ex ) 2. (c) Find the equations of all horizontal asymptotes of h, or explain why none exist. Justify using it computations. (5 points) Our computations below are aided by making the substitution t = e x ; notice that x t = and x t = 0. h(x) x t et 2 = 0 2 = 2 h(x) x t 0 et 2 = e 0 2 = 1 2 = 1 So horizontal asymptotes are y = 2 and y = 1. (d) It is a fact that h is one-to-one (which you do not have to prove). Find an expression for h 1 (x), the inverse of h. (4 points) so h 1 (x) = ln( ln(x + 2)). y = e ( ex) 2 y + 2 = e ( ex ) ln(y + 2) = e x ln(y + 2) = e x ln( ln(y + 2)) = x

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 6 of 11 4. (9 points) The day length T (in minutes) on October 15th varies with latitude L (in degrees North of the Equator) according to the following table: L 30 35 40 45 50 T (L) 703 699 695 690 684 (a) Give your best estimate for the value of T (40), showing your reasoning, and make sure to specify the units of this quantity. (3 points) By definition, T (40) L 40 T (L) T (40). L 40 Since only discrete values of T (L) are given, our best estimate is found by averaging the values of this difference quotient for the values of L nearest to 40, namely for L = 35 and L = 45. Thus, T (40) 1 [ ] T (35) T (40) T (45) T (40) + 2 35 40 45 40 = 1 ( ) 699 695 690 695 + 2 5 5 = 9 minutes of day length on Oct. 15th. 10 degree north of the equator (b) What is the practical meaning of the quantity T (40)? Give a brief but complete one- or twosentence explanation that is understandable to someone who is not familiar with calculus. (3 points) T (40) is the instantaneous rate of change of day length on Oct. 15th in minutes with respect to latitude in degrees north of the equator, at a latitude of 40 degrees north. In other words, when the latitude changes slightly from 40 degrees north, the minutes of day length on Oct. 15th changes by about 9/10 per degree changed. (c) The average latitude of the U.S. is 38 degrees North. Apply your value of T (40) to approximate the day length on October 15th at this latitude, showing your reasoning. (3 points) At 40 degrees north, the day length on Oct. 15th is T (40) = 695 minutes. Since 38 40 = 2 and each change in degree corresponds to a change of approximately 9/10 minutes by (b), we reason that ( T (38) 695 + 9 ) ( 2) = 695 + 9 = 696.8 minutes. 10 5 Note that this is sensible because it is between T (35) = 699 and T (40) = 695. Remark. Based on what we now know about linear approximation, we see that this is just T (38) T (40) + T (40)(38 40).

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 7 of 11 5. (8 points) Let f(x) = 3x + 2. Find f (x) using the it definition of the derivative. Show the steps of your computation. f f(x + h) f(x) (x) h 0 h 3(x + h) + 2 3x + 2 h 0 h ( 3(x + h) + 2 3x + 2)( 3(x + h) + 2 + 3x + 2) h 0 ( 3(x + h) + 2 + 3x + 2) h 3(x + h) + 2 (3x + 2) h 0 ( 3(x + h) + 2 + 3x + 2) h 3h h 0 ( 3(x + h) + 2 + 3x + 2) h 3 h 0 3(x + h) + 2 + 3x + 2) = 3 2 3x + 2

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 8 of 11 6. (8 points) Find the derivative, using any method you like. You do not need to simplify your answers. (a) g(x) = x4 4 1 + e + 2 x 3 3 x 2 + π2 (4 points) Therefore, g(x) = 1 1 + e (x4 4) + x3/2 + π2 x2/3 = 1 1 + e (x4 4) + x 3/2 2/3 + π 2 = 1 1 + e (x4 4) + x 5/6 + π 2. g (x) = 1 1 + e (4x3 ) + 5 6 x 1/6. Alternate Solution: Via the quotient rule: g (x) = d (1 + e) dx (x4 4) (x 4 4) d dx (1 + e) (1 + e) 2 + x2/3 d dx (x3/2 ) x 3/2 d dx (x2/3 ) (x 2/3 ) 2 = (1 + e)(4x3 ) (1 + e) 2 + x2/3 ( 3 2 x1/2 ) x 3/2 ( 2 3 x 1/3 ) (x 2/3 ) 2. (b) h(x) = x e e x + 5 tan x cos x + 1 (4 points) Use the product rule and quotient rule: h (x) = x e d dx (ex ) + e x d d d (cos x + 1) dx (xe dx (5 tan x) (5 tan x) dx (cos x + 1) ) + (cos x + 1) 2 = x e e x + e x (ex e 1 ) + (cos x + 1)(5 sec2 x) (5 tan x)( sin x) (cos x + 1) 2

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 9 of 11 7. (5 points) Show that the graph of f(x) = e x cos x + x has a horizontal tangent line at some point (a, f(a)), where a lies in the interval [0, π]. There is a horizontal tangent at (a, f(a)) precisely when f (a) = 0. f (x) = d dx (ex ) cos x + e x d (cos x) + 1 dx = e x cos x e x sin x + 1 f (0) = e 0 cos 0 e 0 sin 0 + 1 = 1(1) 1(0) + 1 = 2 > 0; f (π) = e π cos π e π sin π + 1 = e π ( 1) e π (0) + 1 = e π + 1 < 0, because π > 0 means e π > e 0 = 1 (e x is an increasing function.) f (0) > 0 > f (π) and f is a continuous function, so, by the Intermediate Value Theorem, there is an a in [0, π] where f (a) = 0.

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 10 of 11 ax 2 + b if x < 4 8. (8 points) For constants a, b, and c, let f(x) = c if x = 4 x if x > 4 (a) Find, with complete reasoning, necessary conditions on a, b, c so that x 4 f(x) exists. (3 points) In order for f to have a it at x = 4, we need f(x) = x 4 expression for f, we have f(x) + b) = 16a + b. x 4 x 4 (ax2 On the other hand, f(x) x = 2. Hence, the only condition we need is x 4 + x 4 + Remark. Note that c can take any real value. 16a + b = 2. f(x). Given the x 4 + (b) Find, with complete reasoning, necessary conditions on a, b, c so that f is differentiable at x = 4. (5 points) First note that if f is differentiable at x = 4, then f must also be continuous at x = 4. In order for f to be continuous at x = 4, we need x 4 x 4 f(x) f(x) = f(4), + which gives us that 16a + b = 2 = c. Now, by definition, f is differentiable at x = 4 if the following holds: f(x) f(4) f(x) f(4). x 4 x 4 x 4 + x 4 For the left-hand it, we have On the other hand f(x) f(4) ax 2 + b c x 4 x 4 x 4 x 4 ax 2 16a (Since c = 2 and b = 2 16a) x 4 x 4 a(x 4)(x + 4) x 4 x 4 a(x + 4) x 4 = 8a. f(x) f(4) x 2 x 4 + x 4 x 4 + x 4 x 4 x 4 + (x 4)( x + 2) 1 x 4 + x + 2 = 1 4. This gives us 8a = 1 4, so a = 1 32, and it follows that b = 3 2. Finally, recall c = 2.

Math 41, Autumn 2013 Solutions to First Exam October 15, 2013 Page 11 of 11 9. (12 points) (a) On the set of axes below, sketch the graph of a function f that satisfies all of the following conditions. Be sure to label the scales on your axes. f is continuous on (, ) f( 2) = 0 and f( 1) = 1 f (x) = 1 for x < 2 f( x) = f(x) for all x satisfying 1 x 1 x 0 f (x) = f(x) > 0 for x > 0 the only local maximum of f(x) for x > 0 is located at x = 1 the only inflection point of f(x) for x > 0 is located at x = 2 (6 points) Note that the above conditions imply that f must pass through the origin and have a vertical tangent here. However, the sketch below represents only one possible graph satisfying all the conditions. (In particular, there are several possible shapes for f over the intervals [ 2, 1] and [2, ). The corner at x = 2 is not necessary, for example.) (b) On the axes below, sketch the graph of f (the derivative of f), labeling your axes as appropriate. (6 points) This must be consistent with part (a) and the conditions above. In particular, note that f must be even on [ 1, 1]; and f must have exactly one turning point over the postive x-axis (namely, at x = 2).