Contents. 6 Graph Sketching Increasing Functions and Decreasing Functions Intervals Monotonically Increasing or Decreasing...

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1 Contents 6 Graph Sketching Increasing Functions and Decreasing Functions Intervals Monotonically Increasing or Decreasing Etrema Maima and Minima Relative Maima and Relative Minima The Second Derivative Test for Relative Etrema Concavity Points of Inflection Asymptotes Symmetry Graph Sketching Graphs of Trancendental Functions Polynomials Functions with Parameters

2 2 CONTENTS

3 Chapter 6 Graph Sketching 6.1 Increasing Functions and Decreasing Functions For problems numbered 1 to 5 show whether the function is increasing or decreasing at the indicated points. 1. (a) f() = 2 2 1; at = 0, = 3, and = 1 2 (b) f() = ; at = 1, = 1, = 2, and = 4 (c) f() = 2; at = 2, = 1, and = 2 (d) f() = 1 ; at = 0, = 1 2, and = 2 2. (a) f() = ; at = 1, = 0, and = 3 (b) f() = +1 ; at = 1, and = 10 (c) f() = ; at = 2, = 1, = 0, and = 1 (d) f() = (2+1) ( 2) ; at = 7 (e) f() = 2 +1 ; at = 1 3. (a) h() = cos 2 ; at = π 4 (b) g() = sin 2; at = 0, = π 4 and = 3π 4 (c) g() = tan ; at = π 4, = 0, and = π 4 (d) g() = sin ; at = π 2, and = 0 4. (a) f() = e ; at = 10, = 1, = 0, and = 1 (b) f() = ( + 1)e ; at = 10, = 2, = 1, and = 0 (c) f() = e ; at = 1, = 1, and = 10 (d) f() = e ( 1)2 ; at = 1, = 1, and = 2 5. (a) h() = ln(1 ); at = 2, and = 0 (b) h() = ln ; at = 1, = e 2, and = e 3 (c) h() = ln 2 ; at = e 1, = 1, and = e (d) h() = ln(sin ); at = π 4, = π 2, and = 3π 4

4 88 Graph Sketching 6.2 Intervals Monotonically Increasing or Decreasing For problems numbered 6 to 11 divide the domain of the function into a finite number of intervals on each of which the function is strictly monotone. Indicate the intervals where the function is increasing and the intervals where it is decreasing. Eample: f() = Solution: f () = 12( ) = 12( + 1)(2 3)( 2) (see Chapter 0 page 5 Theorem I) f () = 0 at = 1, 3 2 or 2. Because f is a continuous function we can conclude f() is strictly monotone on each of the intervals (, 1), ( 1, 3 2 ), ( 3 2, 2), (2, + ). Compute a value f ( 2) = 336 < 0 to conclude that f is decreasing on (, 1) due to the fact f is continuous on (, 1) and 2 (, 1). f (0) > 0 increasing on ( ) 1, 3 2 f ( ) ( 7 4 < 0 decreasing on 3 2, 2) f (3) > 0 increasing on (2, + ). NOTE: To use this method you must check that f is continuous on the given interval and state this as part of your solution. 6. (a) f() = ( 1) (b) f() = (c) f() = 2 (d) f() = 7. (a) f() = 1 ( 1)( 2) (b) f() = (c) f() = (2 +1) 2 (d) f() = 3 8. (a) f() = ( ) 1 2 (b) f() = 1 ( 2 +4) 1 2 (c) f() = ( 1)( 2)( 3) (d) f() = 2 9. (a) f() = e (b) f() = e 2 (c) f() = e +1 (d) f() = e (a) f() = 2 ln (b) f() = ln (c) f() = ln (d) f() = ln( 2 + 1) 11. (a) s() = sin ( π 2 ) (b) f() = sin (c) f() = sin (d) f() = sin + cos

5 6.3 Etrema Maima and Minima Etrema Maima and Minima Eamples maimum maimum relative maimum relative maimum relative minimum minimum minimum relative minimum a b The maimum (global maimum) is the highest value a function attains on the given domain. The minimum (global minimum) is the lowest value a function attains on the given domain. Some functions do not have a maimum or a minimum. For each of the following problems find the global maimum and the global minimum. If it eists, state the place and value. If it does not eist, state why not. Domain is (, + ) unless otherwise stated. 12. f() = f() = f() = 15. f() = sin on [0, 2π] 16. f() = cos on [0, 2π] 17. f() = 2 on f() = 2 on < f() = tan 20. f() = csc 21. f() = sec 22. f() = 2 on ( 2, +2) 23. f() = 2 on > f() = 1 on ( 1, 3) 25. f() = on ( 3, 17] 26. f() = 2, for > f() = 5 2 on > f() = f() = f() = ( 1)( 2)( 3) 31. f() = sin 32. f() = ( 1)( 2)( 3) 33. f() = f() = f() = 2 + 1, f() = 37. f() = on < 3 This means domain of is (, 4) and (4, )

6 90 Graph Sketching 38. f() =, f() < f() = 25 2 on < f() = 25 2 on < f() = sin 42. f() = 2 on ( 2, +2) 43. f() = 2 on [ 2, +2] 44. f() = 3 on [ 2, +2] 45. f() = 46. f() = 2 on ( 2, +2) 47. f() = 2 on [ 2, 2] 48. f() = 49. f() = 2 on ( 2, +2) 50. f() = 2 on [ 2, +2] 51. f() = f() = ( 5) f() = 2 ( 5) f() = f() = sin on > on 3 < 0 on f() = 1 on < 1 on ( 1, +1] 1 on > f() = tan 58. f() = 2, f() < f() = 2, f() > f() = 2, f() > f() = 2 + 1, f() > f() = f() = s = 128t t y = f(θ) = 1 cos 2 θ on ( π 2, + ) π y = ( 5) 3 on ( 0, 10] 6.4 Relative Maima and Relative Minima Recall that a critical number for a function f is a number c such that either f (c) = 0 or f (c) does not eist. The function f is given by the formula. The range is those values which are less than 9. In other words for each point on the graph, both statements must hold.

7 6.4 Relative Maima and Relative Minima 91 (i) (ii) (iii) (iv) f (c) = 0 rel.ma. f (c) = 0 rel. min. f (c) = 0 no ma. or min. f (c) = 0 no ma. or min. (v) (vi) (point cusp) f (c) does not eist rel. ma. (corner cusp) f (c) does not eist rel.ma. Notice that sometimes a relative maimum or a relative minimum occurs when the derivative does not eist (as in (v) and (vi)). At a point or a corner, the first derivative test still works. For questions numbered 68 to 117 (a) Find the domain of the given function. (b) Find the first derivative of the function. (c) Find the critical numbers. (d) On what intervals is f () > 0, and on what intervals is f () < 0? (e) Use the first derivative test to give the relative maima and the relative minima. (f) If they eist, give global maimum, global minimum, and their location. 68. f() = f() = f() = ( 1) f() = f() = f() = 2 e 74. f() = f() = f() = f() = sin π 78. f() = sin 2 π 79. f() = 2 ( 1) f() = 3 ( 1) f() = ae b2 2, a > 0

8 92 Graph Sketching 82. f() = f() = 1 + 2( 1) f() = 1 + ( 1) f() = 1 3 ( 2) f(t) = e t sin t, on [0, π] 87. f() = f() = f() = f() = f() = (2 )3 2(1 ) ( 92. f() = ln ) 4 (3 4) f() = + sin 94. f() = sin 95. f() = sec tan 96. f() = ln 97. f() = e ln(2 +3) 98. y = y = y = y = y = y = y = y = f() = f() = f() = f() = f() = f() = f() = f() = 1 ( ) 114. y = f() = f() = f() = For each of the questions numbered , sketch a graph of some function with the given properties. (You do not have to give a formula for the function. Just give the picture.) 118. A function with a relative minimum at 2 but no global minimum A function with the domain [0, 2] and a maimum at A function with the domain (0, 2] with a relative maimum at 1 and a maimum at 2. Is f (2) = 0? 121. A function with a relative minimum at 0, but with f (0) not eisting A differentiable function with neither a maimum nor a minimum A function with a relative maimum and a relative minimum but with no maimum or minimum A function whose range is all values less than 1 that does not have a maimum A function with a maimum at 5 but f (5) does not eist.

9 6.5 The Second Derivative Test for Relative Etrema A function which is defined everywhere but has no maimum, no minimum, no relative maimum and no relative minimum A function with a maimum at 5, a minimum at 0, a relative maimum at 2 and a relative minimum at A function with a domain of (0, 2) with a maimum and a minimum A function with a domain of (0, 2) with a relative maimum and a relative minimum A function with a domain of (0, 2) with no maimum, no minimum, no relative maimum and no relative minimum A function whose domain is (, ) and whose range is (π, 5]. Is there a maimum? Is there a minimum? 6.5 The Second Derivative Test for Relative Etrema For questions , use the second derivative test in finding the relative etrema of the indicated functions. If the value of the second derivative is zero at a critical point, use the first derivative test. Then find the etrema f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = ( + 2)( 2) f() = f() = f() = f() = f() = f() = 8 2, f() = 2 3 ( + 2) f() = 2 5 +, f() = 2 3 2, f() = 1 3 ( + 2) f() = + sin 160. f() = f() = f() = + 1, f() = cos

10 94 Graph Sketching 164. f() = 1 9, > f() = f() = f() = f() = f() = (2a ) 1 2, a > f() = (2a 2 ) f() = 3 2 ( 18) Concavity For problems numbered 172 to 189 decide: (a) Is the given function increasing or decreasing at the indicated value of? Why? (b) Is the given function concave upward or concave downward for the indicated value of? Why? 172. f() = , = 3 [decreasing, concave upward] 173. f() = , = f() = , = f() = , = y = , = y = , = y = 4 14, = y = 3 + 9, = y = +1, = 0 [not defined at = 0] 181. y = , = y = cos, = π y = tan sec, = π y = e 2, = y = e, = y = 2 e, = f() = ln ( + 4), = f() = ( + 4)e +4, = f() = ln , = 1

11 6.7 Points of Inflection Points of Inflection In problems numbered 190 to 213 find: (a) Intervals where the function is increasing and decreasing. (b) Relative maimum and relative minimum. (c) Intervals where function is concave up, concave down. Why? (d) Points of inflection. Show why y = 3 [concave up > 0; concave down < 0; inflection at (0, 0)] 191. y = y = y = f() = 1 [concave up > 0, concave down < 0] 195. f() = g() = h() = 3 + a + b 198. y() = sin π [concave up, 1 + 2k < < 2k; concave down 2k < < 1 + 2k; inflection points (k, 0), k Z] 199. T () = tan 200. f() = f() = n, n an integer > φ() = [concave up everywhere] 203. ψ() = 3 + ( + 1) λ() = 3 + ( + 1) c() = sin [concave up everywhere ecept = nπ, n Z] 206. f() = ln f() = e 208. g() = e y = y = f() = + e 212. f() = ln(sin 2 ) 213. g() = sin + 1 sin

12 96 Graph Sketching 6.8 Asymptotes For the functions numbered 214 to 241, (a) Find all the asymptotes if there are any. (b) Does the curve approach the asymptote (i) from below? (ii) from above? (iii) neither? (c) In the case of vertical asymptotes does the curve approach the asymptote at + or? Also check the left and right hand sides of the asymptote separately (a) y = +1 (+2)(+1)( 3) (b) y = 2 + (+2)( 4) 215. (a) y = (b) y = (a) y = 1 ( 2) 2 (b) y = 1 ( 2) (a) y = 1 (b) y = (a) y = (b) y = (a) y = +2 2 (b) y = (a) y = (b) y = (a) y = a 3 (b) y = a+b c d 222. (a) y = (b) y = (a) y = 2 +1 (b) y = (a) y = 2 2 (b) y = (a) y = ( 3)2 (b) y = (a) y = (b) y = (a) y = (b) y = (a) y = (Hint: rationalize the numerator) (b) y = (a) y = ( 1)3 (b) y = 2 ( 2) 2 (+1) 2

13 6.9 Symmetry (a) y = ( ) 3 (b) y = ( ) (a) y = 1 sin (b) y = 1 2 cos 232. (a) y = e (b) y = 1 e 233. (a) y = 2 e (b) y = 2 e 234. (a) y = 1 2 e (b) y = 1 e (a) y = e (b) y = e (a) y = e (b) y = e (a) y = e (b) y = e (a) y = e sin (b) y = e cos 239. (a) y = ln ln +1 (b) y = 240. (a) y = 1 + ln 2 (b) y = e ln 241. Under certain conditions the size of a colony of bacteria is related to time by 2 = t + 1 t + 1 (t 0) where is the number of bacteria and t is the time in hours. To what number does the size of the colony tend over a long period of time? In how many hours will the colony have achieved 95% of its ultimate size? 6.9 Symmetry Symmetry is very easy to test for. It does not do anything spectacular for the graph as does the discovery of asymptotes. However noticing symmetry can shorten your work appreciably An even function, f, is one for which f( ) = f(); and odd function is one for which f( ) = f() for all values in its domain. What can we say concerning the symmetry of the graph of an even function? Of an odd function? 243. Show that if every in an equation is replaced by 2k and the resulting equation is equivalent to the original, then its graph is symmetric about the line = k. For the equations numbered 244 to 263 test for symmetry and find the ais or point of symmetry if one eists (a) y 2 = a 2 2 (b) y = (a) y = 4 (b) y = (a) y 2 = 4 (b) y = (a) y = (b) y = + 1

14 98 Graph Sketching 248. (a) y = 5 2 (b) y = (a) y = a 1 3 (b) y = (a) y = 2 1 (b) y = 251. (a) y = (b) y = 252. (a) y = sin (b) y = cos 253. (a) y = tan (b) y = sec 254. (a) y = csc (b) y = cot 255. (a) y = sin (b) y = sin 256. (a) y = cos (b) y = cos 257. (a) y = sin cos (b) y = 2 sin 258. (a) y = cos (b) y = sin (a) y = 10 2 (b) y = e (a) y = 2e 2 (b) y = e (a) y = e e 2 (b) y = e +e (a) y 2 = 2 3 (b) y 2 = ( 1) 2 ( + 2) 263. (a) y 2 = 2 (1 2 ) (b) y = sin 2 + cos If f() is an even function, is f () even or odd? 265. If f() is an odd function, is f () even or odd? 266. If f is an odd function and g is an even function, is f g even or odd? 267. If f is an odd function and g is an even function is f g even or odd? Is g f even or odd? For each of the following graphs give the following information if the function has the stated property: (a) Domain of the function. (b) Symmetry. (c) For what values of is f() = 0? For what values of y is f(0) = y? (d) On what intervals is f increasing? On what intervals is f decreasing? (e) For what values of is f() a relative maimum or a relative minimum? (f) Estimate the value of where the points of inflection occur. Is the slope of the tangent at the point of inflection equal to zero, greater than zero or less than zero? (g) On what intervals is f concave up and on what intervals is f concave down? (h) Find all asymptotes.

15 6.9 Symmetry Y 269. Y X X 270. Y 271. Y X X 272. Y 273. Y X X

16 100 Graph Sketching 274. Y 275. Y X X 276. Y X 6.10 Graph Sketching In problems 277 to 303 sketch the graph of a function with the listed properties. following information as you can: State as much of the Where is the function increasing, decreasing, concave up, and concave down? Where are the local maima, local minima, and points of inflection? What are the asymptotes and how does the function approach them (e.g., from above or below etc.)? Use one page per problem and at least half the page for your sketch of the function. Do not decide on a scale for your aes until you have come up with a picture of the function. You do not have to use the same scale for both the and y aes. The use of HB pencil and an eraser will make your work easier.

17 6.10 Graph Sketching 101 DOMAIN f f f ASYMPTOTES 277. R {1} f(0) = 0 f < 0 : < < 1 or f > 0 : < < 1 lim 1 < < f < 0 : 1 < < lim 1 + lim f() = R {0} f > 0 : < < 1 or f > 0 : < < 0 y = < < f < 0 : 0 < < lim f() = 0 + f (1) = f ( 1) = 0 lim f() = 0 f < 0 : 1 < < 0 or 0 < < R {0} f(2) = 0 f > 0 : 5 < < 0 or f > 0 : 9 < < 0 y = 2 0 < < f ( 9) = 0; f( 9) = 3 2 lim f() = 0 + f ( 5) = 0; f( 5) = 1 f < 0 : < < 9 or lim 0 f < 0 : < < 5 0 < < 280. R f( 3) = 0 f > 0 : 2 < < f > 0 : < < 1 none f(0) = 0 f ( 2) = 0; f( 2) = 5 f ( 1) = 0; f( 1) = 1 f < 0 : < < 2 f < 0 : 1 < < 281. R f( 3) = 0 f > 0 : < < 2 f > 0 : 2 < < none 3 f( 1) = 0 f > 0 : 2 < < f ( 2 ) = f(2) = 0 f ( 2) = f ( 2 ) = 0 f < 0 : < < f(0) = 6 f < 0 : 2 < < R f( 5) = 0 f > 0 : 2 < < f > 0 : < < 0 lim f(0) = 0 f ( 2) = 0; f( 2) = 3 f (0) = 0 lim f < 0 : < < 2 f < 0 : 0 < < 283. R f(0) = 0 f > 0 : 3 < < 3 f > 0 : 5 < < lim f() = 0 f( 3) = 3 f ( 3) = f (3) = 0 f > 0 : 5 < < 0 lim f() = 0 f(3) = 3 f < 0 : < < 3 f ( 5) = f (5) = 0 = f (0) f < 0 : 3 < < f < 0 : 0 < < 5 f < 0 : < < f(0) = 2 f > 0 : 0 < < 2 f < 0 : < < 2 lim and f > 0 : 2 < < f < 0 : 2 < < lim 2 f (0) = 0 f > 0 : 2 < < 2 = 2 f < 0 : 2 < < 0 = 2 f() = 1 f < 0 : < R f(0) = 1 f > 0 : < < 3 f > 0 : < < 4 lim f() = 0 2 f() = 0 f( 3) = 4 f > 0 : 0 < < 3 f > 0 : 2 < < 2 lim f(3) = 4 f ( 3) = f (3) = f (0) = 0 f > 0 : 4 < < f < 0 : 3 < < 0 f ( 4) = f ( 2) = f (2) = f (4) = 0 f < 0 : 3 < < f < 0 : 4 < < 2 f < 0 : 2 < < 4

18 102 Graph Sketching DOMAIN f f f ASYMPTOTES 286. R f(0) = 1 f > 0 : 0 < < f > 0 : < < 0 none f() = f( ) f (0) is undefined f > 0 : < < f < 0 : < < 0 f (0) is undefined 287. R f(0) = 30 f > 0 : < < 1 f > 0 : 1 < < 3 none 3 2 f( 1) = 0 f > 0 : 3 < < f > 0 : 3 < < 4 f(3) = 0 f ( 1) = f ( 3 ) = 0 f ( 1 ) = f ( 3 ) = f (3) = f < 0 : 1 < < 3 4 f < 0 : < < 1 3 f < 0 : 3 < < > 1 f(0) = 0 f > 0 : 1 < < 5 2 f > 0 : 1 < < 2 lim 1 + f(2) = 0 f (1) = f ( 5 ) = 0 2 f > 0 : 3 < < lim f < 0 : 1 < < 1 f (2) = f (3) = 0 f < 0 : 5 < < f < 0 : 2 < < R f(0) = 1 f > 0 : 0 < < f > 0 : 1 < < 1 lim f(1) = 0 f (0) = 0 f ( 1) = f (1) = 0 lim f() = f( ) f < 0 : < < 0 f < 0 : < < 1 f < 0 : 1 < < 290. R f(0) = 5 f > 0 : 1 < < f > 0 : 1 < < 2 3 lim f() = 2 f(3) = 0 f (1) = 0 f ( 1 3 ) = f (2) = 0 lim f() = 2 f < 0 : < < 1 f < 0 : < < 1 3 f < 0 : 2 < < 291. R { 1, 1} f(0) = 0 f > 0 : < < 2 f > 0 : < < 1 lim (f() ) = 0 f(2) = 5 f > 0 : 2 < < f > 0 : 0 < < 1 lim f() = 1 + f( 2) = 5 f (2) = f ( 2) = 0 f (0) = 0 lim f() = 1 + f < 0 : 2 < < 1 f < 0 : 1 < < 0 f < 0 : 1 < < 1 f < 0 : 1 < < f < 0 : 1 < < (a) g () eists for every R (b) g() = g( ), R (c) lim g() = 1 (d) g(0) = 0 (e) g () > 0 on 0 < < e and g () < 0 on e < < 293. (a) f is symmetric about the y ais and the domain of f is R (b) (c) lim f() = and lim f() = 0 f () > 0 on 0 < <

19 6.10 Graph Sketching (d) f(1) = 0 (e) f (2) = (a) f is symmetric about the origin and f() is defined for 0 (b) (c) (d) lim f() = 0 and lim f () = f() > 0 on 0 < < f () < 0 on 0 < < 295. (a) Domain of f is (, 5) ( 5, ) (b) f(0) = 0 (c) (d) lim f() = 1 and lim f() = 0 lim f() = and lim f() = (e) f () > 0 on (, 5) and f () < 0 on ( 5, ) 296. (a) Domain of f is R (b) f(3) = 4 (c) f () = 2 3, R 297. (a) Domain of f is [0, 9] (b) The point, (3, 2) is on the graph of f (c) f () = 1 3, [0, 9] 298. (a) Domain of f is R (b) f(3) = 2 and f(7) = 14 3 (c) f () = c, R and c is a constant (a) Domain of f is R (b) (c) f( 1) = 2 and f(3) = 6 f () = 0, R 300. (a) g(3) = 0 (b) g () = 1 2 on (3, ) (c) g () = 1 2 on (, 3)

20 104 Graph Sketching 301. (a) f (0) = 2 and f( 2) = 0 (b) f () = 0 on (, 0] (c) f () = 2 on [0, ) 302. (a) f is a continuous function on R (b) f(0) = 0 and f( 4) = 0 (c) lim f() = and lim f() = (d) f () > 0 on (, 2) (0, ) (e) f () < 0 on ( 2, 0) (f) f () < 0 on (, 3 2 ) and f () > 0 on ( 3 2, ) (g) lim f () = and lim f () = (a) f is symmetric about the y ais (b) (c) lim f() = 3 lim f() = and lim f() = (d) f(0) = 3 (e) f (0) = 0 Eample: Graph f() = Write f() = = +1 (+2). Justification for following chart is: Domain of f is R {0, 2}. + 1 > 0 = > 1, + 2 > 0 = > 2, and > ( + 2) ( + 1) f < 0 f > 0 f < 0 f > lim = lim = 0 implies ais is a horizontal asymptote. The above chart shows how the + 2 function approaches the ais. Vertical asymptotes (you must compute these limits!) lim f() = 2 lim f() = 2 + lim f() = lim f() = 0 0 +

21 6.10 Graph Sketching 105 Intercepts: + 1 = 0 = 1 is the intercept. f () = ( + 1)(2 + 2) ( 2 + 2) 2 = ( ) ( 2 + 2) 2 = ( + 1)2 + 1 ( 2 + 2) 2 < 0 for all real numbers; thus f is decreasing on its domain. f () = (2 + 2)(2 + 2) 2 2( )( 2 + 2)(2 + 2) ( 2 + 2) 4 f () = 22 ( + 1)( + 2) 2 4( )( + 2)( + 1) 4 ( + 2) 4 f () = 2( + 1)( + 2){( + 2) 2( )} 4 ( + 2) 4 f () = 2( + 1){( + 1)2 + 3} 3 ( + 2) 3 ( + 2) 3 > > 0 > 2, 3 > 0 > 0 and + 1 > 0 > 1 provides information required for this chart: ( + 1) ( + 2) f < 0 f > 0 f < 0 f > 0 concave concave concave concave down up down up Inflection point occurs at = 1 f() 2 1 Eample: Graph f() = ( 3) 2/3 ( + 1). Domain of f is R because cube root function is always defined. Intercepts f(0) = ( 3) 2/3 = 9 1/3, f( 1) = 0, and f(3) = 0. Asymptotes: lim f() = and lim f() =. There are no asymptotes.

22 106 Graph Sketching Increasing / decreasing: f () = 2( + 1) 3( 3) 1/3 + ( 3)2/3 = 5 7 3( 3) 1/3. Since 5 7 > 0 = > 7/5 and 3 > 0 = > 3 we can construct the following chart: /5 3 f increasing r decreasing r increasing (, 7/5) (7/5, 3) (3, ) ma/min By the First Derivative Test we conclude f has a local maimum at = 7/5 and a local minimum at 3 where f is undefined. Concavity: f () = 5 ( 3( 3) 1/3) 3(5 7) 1 3 ( 3) 2/3 9( 3) 2/3 [ ] ( 3) 2/3 ( 3) 2/3 f () = 15( 3) (5 7) 2(5 19) = 9( 3) 4/3 9( 3) 4/3 f () > 0 iff 5 19 > 0 = > Concave up on the interval ( 19 5, ). Concave down on the interval (, 19 5 ). Inflection points: Since f(19/5) is defined and concavity changes at the point (19/2, f(19/2)), it is the only inflection point

23 6.10 Graph Sketching 107 y (19/5, f(19/5)) 1 7/5 3 Eample: g() = 1/ Domain of g is ( 1, ) because > 0 is needed for square root to be defined in the denominator of g lim g() = 0 because increases without bound. lim g() = because numerator remains nonzero as denominator 1 + approaches zero through positive values. ( ) 3 2 g 2 1+ () = (1 + 3 = ) 2(1 + 3 ). 3/2 g is always decreasing on ( 1, ) since g () < 0 on domain of g. (1 + 3 ) > 0 and 2 0 ecept at = 0. There are NO maimum or minimum points. [ 2(1 + 3 ) 3/2 3 4 (1 + 3 ) 1/2 ] g () = 3 2 (1 + 3 ) 3 g () = ( 3 2(1 + 3 ) 3 3) 2(1 + 3 ) 3 = (2 3 ) 2(1 + 3 ) 3. ( 3) (2 3 ) /3 g > 0 g < 0 g > 0

24 108 Graph Sketching Conclusion: g is concave up on ( 1, 0) and (2 1/3, ) and concave down on (0, 2 1/3 ). There is a change of concavity at = 0 and = 2 1/3. Also g is defined at these two points so they qualify as inflection points. g() For each function or relation given in problems numbered 304 to 373, sketch the graph. Determine (a) The domain. (b) Any symmetry, if it eists. State why. (c) The asymptotic behaviour. State why. (d) The intervals where the function or relation is increasing, decreasing. State why. (e) The relative maima, relative minima, the maimum and the minimum if they eist. State why. (f) The intervals where the function or relation is concave up, concave down. State why. (g) The points of inflection. State why. Include the above information on the graph y = y = y = f(r) = E2 R r+r E > 0, r > 0 are constants 308. y = y = y = y =

25 6.10 Graph Sketching f() = y = f() = +a +a 315. g() = y = h() = y = cos ( θ + π 4 ) 319. y = sin f(θ) = sin θ 321. h() = 4 sin ( ) f(θ) = 1 + cos θ 321. f() = tan ( ) π 324. f() = f() = g() = y = y = ( 4) (+1)( 2) 329. y = y = ( 1) y = y = f() = (+1)2 (+2) 2 ( 1) 334. f() = f() = f () and f () for the preceding question 337. g() = g (), g (), g () for the preceding question 339. h() = A(z) = z 4 6z 2 + 8z y = 3 2 ( 4) 342. y = y = 4 3 ( 2 4) 344. y = 1 3 ( 4) 345. y = 3 2 ( 2 + 2) 346. y = ( + 2) 3 ( 1) 347. y = y = P (R) = E2 R (r+r) 2 E > 0, r > 0 are constraints 350. f() = f() = 2 sin 2 + cos f() = f() = sin sin 354. f() = 2 tan sec 355. f() = ( 1) 2 4

26 110 Graph Sketching 356. f() = 3 ( + 4) 357. f() = f() = f() = f() = f() = f() = f() = f() = f() = sin, < π 366. f() = π 2, > π 2 { 2 + 1, < f() = + 1, > f() = f() = f() = 2 ( 2 4)( 1) 371. f() = csc f() = ( 1) f() = ( ) 6.11 Graphs of Trancendental Functions For problems numbered 374 to 394 sketch the functions f and g on the same coordinate aes f() = e and g() = f() = e and g() = f() = e and g() = f() = e and g() = e 378. f() = 2 and g() = f() = 10 and g() = f() = 2 and g() = log f() = 3 and g() = log f() = ( 1 2) and g() = log f() = 3 and g() = f() = ln and g() = log f() = ln and g() = ln( + 1) 386. f() = ln and g() = ln( 1) 387. f() = 3 and g() = f() = 10 and g() = f() = ln(2) and g() = ln

27 6.11 Graphs of Trancendental Functions f() = ln ( ) 3 and g() = ln 391. f() = 1 2 and g() = ( 1 2 ) 392. f() = log 2 and g() = log f() = log 2 and g() = log 2 ( + 1) 394. f() = ( 1 3) and g() = log 1 3 For problems numbered 395 to 402, find the equation of the line tangent to the curve at the point. Question numbered 395 is solved as an eample Curve is y = e 2, = 2. Solution: y = 2e 2 + e 2 At = 2, y = 4e 4 + e 4 = 5e 4 At = 2, y = 2e 4 The tangent line has the equation y y 1 = m( 1 ) y 2e 4 = 5e 4 ( 2) or 5e 4 y = 8e 4 Ans Curve is y = 2, = Curve is y = ln, = e 398. Curve is y = 2 e, = Curve is y = ln, = e [2 y e = 0] 400. Curve is y = 2 e 2, = Curve is y = ln 2, = Curve is y = (ln ) 2, = e For problems numbered 403 to 421, find: (a) the intervals in which f is increasing, and those in which f is decreasing; (b) the relative maima and minima; (c) the intervals where the curve is concave upward and where it is concave downward; (d) the inflection points. and sketch the graph f() = f() = 10 (+1) 405. f() = e 406. f() = e f() = e 408. f() = e f() = e e f() = e +e 2

28 112 Graph Sketching 411. f() = e f() = 2 e 413. f() = ln( 2 1) 414. f() = 2e ( 1 2 ) f() = 2 e f() = ln(1 + ) 417. f() = ln 418. f() = 2 2 ln 419. f() = 2(ln ) f() = e sin 421. f() = e cos Sketch the graphs of functions numbered 422 to 456 and provide the following information. (a) The domain. (b) Any symmetry, if it eists. State why. (c) The asymptotic behaviour. State why. (d) The intervals where the function or relation is increasing, decreasing. State why. (e) The relative maima, relative minima, the maimum and minimum if they eist. State why. (f) The intervals where the function or relation is concave up, concave down. State why. (g) The points of inflection. State why. Include all the above information on the graph. In cases where the function or the derivative takes an indeterminate form, find the limit y = e 423. y = 2 ln 424. y = ln 425. y = ln 426. y = e 427. y = e y = e y = 2(ln ) y = ln 431. y = e y = 1 ln 433. y = 1+ln 434. y = e y = ln 436. y = ln 437. y = ln y = ln y = e y = cos e 441. y = e y = e 443. y = ln

29 6.12 Polynomials y = ln 445. y = e 446. y = ln 447. y = + ln 448. y = ln 449. y 2 = ln 450. y 2 = e 451. y 2 = e 452. y 2 = e (8 2 ) 453. y 2 = ln 454. y 2 = ln 455. y 2 = ln 456. For y = e, sketch the graph of y, y and y Polynomials 457. Find polynomials which have the following properties then draw their graphs making estimates whenever necessary. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) no roots, one minimum no roots, one maimum one root, no etremum one root, no etremum one root, one maimum one root, one minimum one root, no etremum; one point of inflection one root, two etremum three roots all distinct three roots all distinct and five etremum two roots, two etremum two roots, three etremum no roots, three etremum 458. Same instructions as in question 458 (a) a root at = 1 and a maimum at = 2 (b) a root at = 1 and a minimum at = 2 (c) roots at = 2, 4 and a maimum at = 3 (d) roots at = 2, 4 and a minimum at = 3 (e) (f) a root at = 1 and two etremum somewhere a root at = 0 and a point of inflection somewhere 459. graph the following polynomials estimating whenever necessary. Include numerical estimates (or eact values if possible) of roots and etreme values.

30 114 Graph Sketching (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) Functions with Parameters For the function or relation given in the following problems do the following: (a) For each problem decide how to partition the range of the parameter. If c is the parameter the three cases: c < 0, c = 0 and c > 0 may be appropriate. Other possibilities are c < 1, c = 1 or just one case c R. (b) For each case compute and describe all the important features of its graph. Draw a neat sketch of the graph and label these features on your diagram: 460. (a) f() = 2 2a + a 2 (b) a (c) f() = 2 + a (d) 2 + a + 3 (e) f() = 2 + a 2 3 (f) ( a) (a) A() = a (b) A() = ( a)2 a 462. (a) l() = (a 2 1) (b) l() = (a + 3) a (c) l() = a + (a 4) 463. (a) p() = ( a) 2 + 2( a) 6 (b) p() = ( a) 3 + ( a) 2 χ + a 464. (a) g() = 1 (a 1) (b) g() = 1 a 465. (a) f() = 4χ 1 a 2 (b) f() = + 1 a 1 +a { 2 a if a 466. g() = a 2 if < a 467. (a) g() = 2 a (b) g() = a (c) g() = 2 a 2 (d) g() = a 468. ( 1) 2 + (y 1) 2 = 2 ln(c) (y c) 2 = c 2