2.5 Continuity. f(x) + g(x) > (M c) + (c - 1) == M. Thus,

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1 96 D CHAPTER LIMITS AND DERIVATIVES If() - LI < c. Let 6 be the smaller of 61 and 6. Then 0 < I - al < 6 =} a - 61 < X < a or a < < a + 6 so If() - LI < c. Hence, lim f() == L. So we have proved that lim f() == L {::> lim f() == L == lim f() ( + 3)4 > 10,000 {::> ( + 3) < 10,000 {::> I + 31 < ~ I - (-3)1 < Given M > 0, we need <5 > 0 such that 0 < I + 31 < <5 =? l/( + 3)4 > M. Now ( ~ 3)4 > M ~ ( + 3) < M ~ I + 31 < {1M" So take <5 = {1M" Then 0 < I + 31 < <5 = {1M =? ( + 3)4 > M, so. 1 1im == ( + 3)4 43. Given M < 0 we need 6 > 0 so that ln z < M whenever 0 < < 6; that is, == e 1n < em whenever 0 < < 6. This suggests that we take 6 == em. If0 < < em, then In < In em == M. B the definition of a limit, lim In == o+ 44. (a) Let M be given. Since lim f() == 00, there eists 61 > 0 such that 0 < I - al < 61 =} f() > M c. Since lim g() == c, there eists 6 > 0 such that 0 < I - al < 6 =} Ig() - c < 1 =} g() > c- 1. Let 6 be the smaller of 61 and 6. Then 0 < I - al < 6 =} f() + g() > (M c) + (c - 1) == M. Thus, lim [f() + g()] == 00. (b) Let M > 0 be given. Since lim g() == c > 0, there eists 61 > 0 such that 0 < I - al < 61 =} Ig() - cl < c/ =} g() > c/. Since lim f() == 00, there eists 6 > 0 such that 0 < I - al < 6 =} f() > M/c. Let 6 == min {61,6}. Then 0 < I - al < 6 =} f() g() > M ~ == M, so lim f() g() == 00. c (c) Let N < 0 be given. Since lim g() == c < 0, there eists 61 > 0 such that 0 < I - al < 61 =} Ig() -cl < -c/ =} g() < c/. Since lim f() == oo.fhere eists o, > osuch that 0 < I-al < 6 =} f( ) > N/ c. (Note that c < 0 and N < 0 =} N/ c > 0.) Let 6 == min {61, 6}. Then 0 < I - al < 6 =} f() > N/c =} f() g() < N. ~ == N, so lim f() g() == -00. c.5 Continuit 1. From Definition 1, lim f() == f(4) The graph of f has no hole, jump, or vertical asmptote. 3. (a) f is discontinuous at -4 since f( -4) is not defined and at -,, and 4 since the limit does not eist (the left and right limits are not the same). (b) f is continuous from the left at - since lim f () == f (- ). f is continuous from the right at and 4 since -+- lim f () == f () and lim f () == f (4). It is continuous from neither side at -4 since f (-4) is undefined is continuous on [-4, -), (-,), [,4), (4,6), and (6,8).

2 SECTION.5 CONTINUITY D The graph of == f () must have a discontinuit at == 3 6. and must show that lim f() == f(3) (a) Cost (in dollars) (b) There are discontinuities at times t == 1,, 3, and 4. A person parking in the lot would want to keep in mind that the charge will jump at the beginning of each hour. o 1 Time (in hours) 8. (a) Continuous; at the location in question, the temperature changes smoothl as time passes, without an instantaneous jumps from one temperature to another. (b) Continuous; the temperature at a specific time changes smoothl as the distance due west from New York Cit increases, without an instantaneous jumps. (c) Discontinuous; as the distance due west from New York Cit increases, the altitude above sea level ma jump from one height to another without going through all of the intermediate values - at a cliff, for eample. (d) Discontinuous; as the distance traveled increases, the cost of the ride jumps in small increments. (e) Discontinuous; when the lights are switched on (or off), the current suddenl changes between 0 and some nonzero value, without passing through all of the intermediate values. This is debatable, though, depending on our definition of current. 9. Since f and 9 are continuous functions, lim [f () - 9()] == lim f () - lim 9( ) [b Limit Laws and 3] == f(3) - g(3) [b continuit of f and 9 at == 3] ==. 5 - g(3) == 10 - g(3) Since it is given that lim [f() - g()] == 4, we have 10 - g(3) == 4, so g(3) == lim f() = lim ( + V7 - ) = lim + Jlim 7 - lim = 4 + V7-4 = 16 + v'3 = f(4) B the definition of continuit, f is continuous at a == lim f()== lim (X+X 3)4== (lim + lim 3 )4= [-1+(-1)3t = (-3)4 =81=f(-1) l l l l B the definition of continuit, f is continuous at a == -1.

3 98 D CHAPTER LIMITS AND DERIVATIVES t - 3t lim(t - 3t ) lim t - 3 lim t t--'1-1 t--' lim h(t) = lim = _t--'1-_1 _ (1) - 3(1) = -1 = h(l) t--'1-1 t--' t 3 lim(l + t 3 ) lim 1 + lim t (1)3. t--'1-1 t--'1-1 t--'1-1 B the definition of continuit, h is continuous at a = For a >, we have 3 lim ( + 3) a lim f() = lim +_ = _X--'1- --'1- a --'1- a X - lim ( - ) lim X + lim 3 lim - lim X--'1-a a +3 a- = f(a) X--'1-a [Limit Law 5] [1,, and 3] [7 and 8] Thus, f is continuous at = a for ever a in (, 00); that is, f is continuous on (, 00). 14. For a < 3, we have lim g() = lim -J3=X = lim -J3=X [Limit Law 3] = /Jim (3 - ) [11] X--'1-a X--'1-a X--'1-a VX--'1-a = Jlim 3 - lim [J = I3="a [7 and 8J = g(a), so 9 is continuous at = a for ever a in (-CXJ, 3). X--'1-a X--'1-a Also, lim g() = 0 = g(3), so 9 is continuous from the left at 3. Thus, 9 is continuous on (-00,3]. 15. f () = In I - 1 is discontinuous at since f () = In 0 is not defined. = o 1) if -I- 1. disconti 1 b 1 f() if = 1 X--' f() -_ {1 / (X - IS iscontmuous at ecause nn does not eist. (1,) o L e if < f() = { if.> 0 The left-hand limit of f at a = 0 is lim f() = lim ex = 1. The X--'1-0- X--'1-3 X--'1-0 right-hand limit of f at a = 0 is lim f() = lim = o. Since these X--'1-0+ X--'1-0+ o limits are not equal, lim f() does not eist and f is discontinuous at o. X--'1-0

4 SECTION.5 CONTINUITY D f () = { ;-1 X - X if -I- 1 if = 1 j' J! =l j f() li - X 11m = 1m-- = Ii 1m ( - 1) = l' 1m-- = 1 -+l (+l)(-l) l ' but f (1) = 1, so f is discontinous at 1. =-l cos if < f() = 0 if = 0 { 1 - if > 0 lim f() = 1, but f(o) = 0 -I- 1, so f is discontinuous at O o -7T o X f() = 6-3 { if -I- 3 7 if = = + 1 1im f() = li im = li rm (+l)(-3) = li im ( + 1) = X X ' but f (3) = 6, so f is discontinuous at F () = X is a rational function. So b Theorem 5 (or Theorem 7), F is continuous at ever number in its domain, { I I- O} = {, ( + 3)( + ) -I- O} = { I -I- -3, - } or (-00, -3) U (-3, -) U (-,00).. B Theorem 7, the root function ~ and the polnomial function are continuous on JR. B part 4 of Theorem 4, the 3 product G() = ~ (1 + ) is continuous on its domain, JR. 3. B Theorem 5, the polnomials and - 1 are continuous on (-00,00). B Theorem 7, the root function VX is continuous on [0,(0). B Theorem 9, the composite function -j - 1 is continuous on its domain, [~, (0). B part 1 of Theorem 4, the sum R() = + / - 1 is continuous on [~, 00). 4. B Theorem 7, the trigonometric function sin and the polnomial function + 1 are continuous on JR. B part 5 of Theorem 4, h() = sin is continuous on its domain, { I -I- -I}. +l 5t 5. B Theorem 7, the eponential function e- and the trigonometric function cos 1ft are continuous on (-00,00). 5t B part 4 of Theorem 4, L(t) = e- cos 1ft is continuous on (-00,00). 6. B Theorem 5, the polnomial - 1 is continuous on (-00, (0). B Theorem 7, sin- 1 is continuous on its domain, [-1, 1]. B Theorem 9, sin-1 ( - 1) is continuous on its domain, which is { 1-1 < -1 < I} = { I0 ~ < } = { Ill < J} = [-J,J].

5 100 D CHAPTER LIMITS AND DERIVATIVES 4 7. B Theorem 5, the polnomial t - 1 is continuous on (-00, (0). B Theorem 7, ln is continuous on its domain, (0, (0). B Theorem 9, In (t 4-1) is continuous on its domain, which is {t It 4-1 > o} = {t It 4 > 1} = {t Iitl > 1} = (-00, -1) U (1, (0) 8. B Theorem 7, VX is continuous on [0,(0). B Theorems 7 and 9, eft is continuous on [0,(0). Also b Theorems 7 and 9, cos (ev'x) is continuous on [0,(0) The function = 1/ IS discontinuous at = 0 because the 1 + e left- and right-hand limits at = 0 are different. -4 ~ ~ ~ The function = tan is discontinuous at = ~ + tck, where k is an integer. The function = In (tan ) is also discontinuous where tan is 0, that is, at = tck: So = In (tan ) is discontinuous at = ~ n, n an integer. 31. Because we are dealing with root functions, 5 + VX is continuous on [0, (0), V + 5 is continuous on [-5, (0), so the quotient f() = 5~ is continuous on [0, (0). Since f is continuous at = 4, lim f() = f(4) = ~. V 5 + X X Because is continuous on IR, sin is continuous on IR, and + sin is continuous on IR, the composite function f () = sin( + sin ) is continuous on IR, so lim f () = f (zr) = sin(jr+ sin Jr) = sin tt = O. X-77[" Because - is continuous on IR, the composite function f () = ex - is continuous on IR, so 1 lim f() = f(1) = e - 1 = eo = Because arctan is a continuous function, we can appl Theorem 8. lim arctan (~ - 4 ) = arctan (lim ( + )( - )) = arctan (lim + ) = arctan ~ ~ X X-7 3( - ) X-7 3. if < f() = { VX if ~ 1 B Theorem 5, since f() equals the polnomial on (-00,1), f is continuous on (-00,1). B Theorem 7, since f() equals the root function VX on (1, 00), f is continuous on (1, 00). At = 1, lim f () = lim = 1 and X-71 - X-71 lim f () = lim VX = 1. Thus, lim f () eists and equals 1. Also, f (1) = -II = 1. Thus, f is continuous at = 1. X-71 + X-71 + X-71 We conclude that f is continuous on (-00,00).

6 SECTION.5 CONTINUITY D 101 sin if < 1f /4 36. f() = { cos if : rr/4 B Theorem 7, the trigonometric functions are continuous. Since f () = sin on (-00, rr/4) and f () = cos on (rr/4, (0), f is continuous on (-00, rr/4) U (rr/4, (0). lim f() = lim sin = sin -i = 1/V"since the sine -----*(7f/4) *(7f/4) function is continuous at rr/4. Similarl, lim f () = lim cos = 1/J b continuit ofthe cosine function at -----*(7f/4) *(7f/4)+ rr/ 4. Thus, lim f () eists and equals 1/V", which agrees with the value f (it/ 4). Therefore, f is continuous at r:/4, so -----*(7f/4) f is continuous on (- 00, (0). I + if :S f () = - if 0 < :S { ( - ) if > (0,1) f is continuous on (-00,0), (0,), and (,00) since it is a polnomial on (,0) each ofthese intervals. Now lim f() = lim (1 + ) = 1 and lim f() = lim ( *o *o *o *o+ ) =, so f is discontinuous at O. Since f(o) = 1, f is continuous from the left at 0. Also, lim f() = lim ( - ) = 0, -----* * lim f() = lim ( - ) = 0, and f() = 0, so f is continuous at. The onl number at which f is discontinuous is O * *+ X + 1 if :S f () = 1/ if 1 < < 3 { v=-3 if : 3 (1,) ~) f is continuous on (-00,1), (1,3), and (3, (0), where it is a polnomial, (1,1) a rational function, and a composite of a root function with a polnomial, (3,0) respectivel. Now lim f() = lim ( + 1) = and lim f() = lim (l/) = 1, so f is discontinuous at *l *l *l *l+ Since f(l) =, f is continuous from the left at 1. Also, lim f() = lim (l/) = 1/3, and -----* *3 lim f () = lim v=-3 = 0 = f (3), so f is discontinuous at 3, but it is continuous from the right at * *3+ X f() = ex { if < 0 if 0 :S :S 1 if > 1 f is continuous on (-00,0) and (1,00) since on each ofthese intervals it is a polnomial; it is continuous on (0,1) since it is an eponential. (0,) (0,1) (1,e) (1,1) Now lim f() = lim ( + ) = and lim f() = lim e" = 1, so f is discontinuous at O. Since f(o) = 1, f is -----*o *o *o *o+ continuous from the right at O. Also lim f() = lim ex = e and lim f() = lim ( *l *l *l *l+ ) = 1, so f is discontinuous at 1. Since f (1) = e, f is continuous from the left at 1.

7 10 0 CHAPTER LIMITS AND DERIVATIVES 40. B Theorem 5, each piece of F is continuous on its domain. We need to check for continuit at r == R. lim F(r) == T-fR- T-fR- T-fR+ T-fR+ r T-fR R R lim GR~r = GR~ and lim F(r) = lim GJ:: = GR~'SO lim F(r) = G~. SinceF(R) = G~, F is continuous at R. Therefore, F is a continuous function of r. e + if < 41. f() == { 3 - e if f is continuous on (-00,) and (,00). Now lim f() == X-f- lim (e + ) == 4e + 4 and X-f lim f () == lim ( 3 - e) == 8 - e. So f is continuous {::} 4e + 4 == 8 - e {::} 6e == 4 {::} e == ~. Thus, for f X-f+ X-f+ to be continuous on (-00,00), e == ~ if < - 4. f() == a - b + 3 if <<3 - a + b if 3 At == : lim f() = lim - 4 lim ( + )( - ) = lim ( + ) = + = 4 lim f() == lim (a - b + 3) == 4a - b + 3 X-f+ X-f+ We must have 4a - b + 3 == 4, or 4a - b == I (1). At == 3: lim f() == lim (a - b + 3) == 9a - 3b + 3 X-f3- X-f- X-f- X - X-f- X - X-f X-f3 lim f () == lim ( - a + b) == 6 - a + b X-f3+ X-f3+ We must have 9a - 3b + 3 == 6 - a + b, or loa - 4b == 3 (). Now solve the sstem of equations b adding - times equation (1) to equation (). -8a+4b== - loa - 4b == 3 a 1 So a == ~. Substituting ~ for a in (1) gives us -b == -1, so b == ~ as well. Thus, for f to be continuous on (-00,00), a==b==~. 4 43: (a) f () = - 1 = ( + 1)( - 1) = ( + 1) ( + 1)( -I) = ( + 1) ( + 1) [or X + 1] -I -I -I for # 1. The discontinuit is removable and 9 () == X + 1 agrees with f for # 1 and is continuous on JR. --) X3_X_ ( (-)(+1) ( 1) [ ] (b) f ( X ) == == == == + or + for #. The discontinuit is removable and 9 () == + agrees with f for # and is continuous on JR. (c) lim f () == lim [sin ] == lim 0 == 0 and lim f () == lim [sin ] == lim (-1) == -1, so lim f () does not X-f'!r- X-f'!r- X-f'!r- X-f'!r+ X-f'!r+ X-f'!r+ X-f'!r eist. The discontinuit at == t: is a jump discontinuit.

8 SECTION.5 CONTINUITY ~----- N= 3 N= o f does not satisf the conclusion of the Intermediate Value Theorem. o f does satisf the conclusion of the Intermediate Value Theorem. 45. f() == + 10 sin is continuous on the interval [31,3], f(31) ~ 957, and f(3) ~ Since 957 < 1000 < 1030, there is a number c in (31,3) such that f(c) == 1000 b the Intermediate Value Theorem. Note: There is also a number c in (-3, -31) such that f(c) == Suppose that f(3) < 6. B the Intermediate Value Theorem applied to the continuous function f on the closed interval [,3], the fact that f() == 8 > 6 and f(3) < 6 implies that there is a number c in (,3) such that f(c) == 6. This contradicts the fact that the onl solutions of the equation f() == 6 are == 1 and == 4. Hence, our supposition that f(3) < 6 was incorrect. It follows that f(3) ~ 6. But f(3) -# 6 because the onl solutions of f() == 6 are == 1 and == 4. Therefore, f(3) > f() == 4 + X - 3 is continuous on the interval [1,]' f(l) == -1, and f() == 15. Since -1 < 0 < 15, there is a number c in (1, ) such that f (c) == 0 b the Intermediate Value Theorem. Thus, there is a root of the equation == 0 in the interval (1,). 48. f() == {IX is continuous on the interval [0,1], f(o) == -1, and f(l) == 1. Since -1 < 0 < 1, there is a number c in (0,1) such that f(c) == 0 b the Intermediate Value Theorem. Thus, there is a root of the equation {IX == 0, or {IX == 1 -, in the interval (0, 1). 49. f() == cos - is continuous on the interval [0,1]' f(o) == 1, and f(l) == cos 1-1 ~ Since < 0 < 1, there is a number c in (0, 1) such that f (c) == 0 b the Intermediate Value Theorem. Thus, there is a root of the equation cos - == 0, or cos ==, in the interval (0,1) f() == In - e- is continuous on the interval [1,], f(l) == _e- ~ -0.37, and f() == In - e- ~ Since < 0 < 0.56, there is a number c in (1,) such that f(c) == 0 b the Intermediate Value Theorem. Thus, there is a root of the equation ln - e- == 0, or ln == e-, in the interval (1,). 51. (a) f() == cos - 3 is continuous on the interval [0,1], f(o) == 1 > 0, and f(l) == cos 1-1 ~ < O. Since 1 > 0 > -0.46, there is a number c in (0,1) such that f(c) == 0 b the Intermediate Value Theorem. Thus, there is a root of the equation cos - 3 == 0, or cos == 3, in the interval (0,1). (b) f(0.86) ~ > 0 and f(0.87) ~ < 0, so there is a root between 0.86 and 0.87, that is, in the interval (0.86,0.87). 5. (a) f() == ln is continuous on the interval [1,]' f(l) == -1 < 0, and f() == In + 1 ~ 1.7 > O. Since -1 < 0 < 1.7, there is a number c in (1,) such that f(c) == 0 b the Intermediate Value Theorem. Thus, there is a root of the equation In == 0, or In == 3 -, in the interval (1,).

9 104 0 CHAPTER LIMITS AND DERIVATIVES (b) f(1.34) ~ < 0 and f(1.35) ~ > 0, so there is a root between 1.34 and 1.35, that is, in the interval (1.34,1.35). I OO 53. (a) Let f() == 100e- / Then f(o) == 100 > 0 and f(100) == 100e- l ~ -63. < O. So b the Intermediate 00 Value Theorem, there is a number c in (0,100) such that f( c) == O. l OO This implies that 100e- e / == 0.01c. (b)using the intersect feature of the graphing device, we find that the root of the equation is == , correct to three decimal places. a 54. (a) Let f() == arctan Then f(o) == -1 < 0 and 3 f(l) == ~ > O. So b the Intermediate Value Theorem, there is a number c in (0,1) such that f(c) == O. This implies that arctanc == 1- c. (b) Using the intersect feature ofthe graphing device, we find that the root ofthe equation is == 0.50, correct to three decimal places. 55. (=}) If f is continuous at a, then b Theorem 8 with 9 (h) == a + h, we have lim f(a + h) == f (lim (a + h)) == f(a). h----+o h----+o 4.5 I--l- =l- = arctan --,AC-~_I I + I4.5-3 (- =) Let e > O. Since lim f(a + h) == f(a), there eists 6 > 0 such that 0 < Ihl < 6 =} h----+o If(a + h) - f(a)1 < c. So if0 < I - al < 6, then If() - f(a)1 == If(a + ( - a)) - f(a)1 < c. Thus, lim f () == f (a) and so f is continuous at a. 56. lim sin (a + h) == lim (sin a cos h + cos a sin h) == lim (sin a cos h) + lim (cos a sin h) h----+o h----+o h----+o h----+o == (lim sin a) (lim cos h) + (lim cos a) (lim sin h) == (sin a) (1) + (cos a) (0) == sin a h----+o h----+o h----+o h----+o 57. As in the previous eercise, we must show that lim cos( a + h) == cos a to prove that the cosine function is continuous. h----+o lim cos (a + h) == lim (cos a cos h - sin a sin h) == lim (cos a cos h) - lim (sin a sin h) h----+o h----+o h----+o h----+o == (lim cos a) (lim cos h) - (lim sin a) (lim sin h) == (cos a) (1) - (sin a) (0) == cos a h----+o h----+o h----+o h----+o 58.. (a) Since f is continuous at a, lim f() == f(a). Thus, using the Constant Multiple Law of Limits, we have lim (cf )() == lim cf () == c lim f () == cf (a) == (cf )(a). Therefore, cf is continuous at a. (b) Since f and 9 are continuous at a, lim f () == f (a) and lim 9 () == 9 (a). Since 9 (a) -I- 0, we can use the Quotient Law.. u ( f) () u f ( ) ~~ f () f (a) ( f) ( ) Th f.. of L inuts: im - == rm -(-) == li () == -(-) == - a. us, - IS contmuous at a. 9 9 X nn 9 gag 9 0 if is rational 59. f () ==... is continuous nowhere. For, given an number a and an 6 > 0, the interval (a - 6, a + 6) { 1 If IS irrational contains both infinitel man rational and infinitel man irrational numbers. Since f(a) == 0 or 1, there are infinitel man numbers with 0 < I - al < 6 and If() - f(a)1 == 1. Thus, lim f() -I- f(a). [In fact, lim f() does not even eist.]

10 SECTION.5 CONTINUITY if is rational 60. g() =... is continuous at O. To see wh, note that -Il ~ g() ~ li, so b the Squeeze Theorem { If IS irrational lim g() = 0 = g(o). But 9 is continuous nowhere else. For if a -# 0 and 8 > 0, the interval (a - 8, a + 8) contains both ----+o infinitel man rational and infinitel man irrational numbers. Since g(a) = 0 or a, there are infinitel man numbers with o< I - al < 8 and Ig() - g(a)1 > lal /. Thus, lim g() -# g(a). 61. If there is such a number, it satisfies the equation = {:} 3 - X + 1 = O. Let the left-hand side ofthis equation be called 1(). Now 1(- ) = - 5 < 0, and 1(-1) = 1 > O. Note also that 1() is a polnomial, and thus continuous. So b the Intermediate Value Theorem, there is a number c between - and -1 such that 1(c) = 0, so that c = c a + 3 b = 0 =?- a( ) + b( ) = O. Let p() denote the left side ofthe last equation. Since p is continuous on [-1, 1], p(-1) = -4a < 0, and p(1) = b > 0, there eists a c in (-1, 1) such that p(c) = 0 b the Intermediate Value Theorem. Note that the onl root of either denominator that is in (-1,1) is (-1 + VS)/ = r, butp(r) = (3VS - 9)a/ -# O. Thus, cisnotarootofeitherdenominator, sop(c) = 0 =? = c is a root ofthe given equation () = 4 sin(1/ ) is continuous on (-00, 0) U (0, (0) since it is the product of a polnomial and a composite of a 4 4 trigonometric function and a rational function. Now since -1 ~ sin(l/) ~ 1, we have _ ~ 4 sin(l/) ~. Because lim (- 4 ) = 0 and lim 4 = 0, the Squeeze Theorem gives us lim ( 4 sin(l/)) = 0, which equals 1(0). Thus, 1 is ----+o ----+o ----+o continuous at 0 and, hence, on (- 00, 00). 64. (a) lim F() = 0 and lim F() = 0, so lim F() = 0, which is F(O), and hence F is continuous at = a if a = O. For ----+o o o a > 0, lim F() = lim = a = F(a). For a < 0, lim F() = lim (-) = -a = F(a). Thus, F is continuous at = a; that is, continuous everwhere. (b) Assume that 1 is continuous on the interval I. Then for a E I, lim 11() I =!lim 1() I = 11(a)I b Theorem 8. (If a is an endpoint of I, use the appropriate one-sided limit.) So 111 is continuous on I. I if > 0 (c) No, the converse is false. For eample, the function 1() =. - is not continuous at = 0, but 11() I = 1 is { -1 If < 0 continuous on JR. 65. Define u(t) to be the monk's distance from the monaster, as a function oftime, on the first da, and define d(t) to be his distance from the monaster, as a function oftime, on the second da. Let D be the distance from the monaster to the top of the mountain. From the given information we know that u(o) = 0, u(1) = D, d(o) = D and d(1) = O. Now consider the function u - d, which is clearl continuous. We calculate that (u - d)(0) = - D and (u - d) (1) = D. So b the Intermediate Value Theorem, there must be some time to between 0 and 1 such that (u - d)(to) = 0 {:} u(to) = d(to). So at time to after 7:00 AM, the monk will be at the same place on both das.