Calculus: Early Transcendentals
Kathleen Miranda
Language: English
Pages: 1113
ISBN: 1429254335
Format: PDF / Kindle (mobi) / ePub
Instructors: click here to access the instructor resources.
Michael Sullivan and Kathleen Miranda have written a contemporary calculus textbook that instructors will respect and students can use. Consistent in its use of language and notation, Sullivan/Miranda’s Calculus offers clear and precise mathematics at an appropriate level of rigor. The authors help students learn calculus conceptually, while also emphasizing computational and problem-solving skills. The book contains a wide array of problems including engaging challenge problems and applied exercises that model the physical sciences, life sciences, economics, and other disciplines. Algebra-weak students will benefit from marginal annotations that help strengthen algebraic understanding, the many references to review material, and extensive practice exercises. Strong media offerings include interactive figures and online homework. Sullivan/Miranda’s Calculus has been built with today’s instructors and students in mind.
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The inverse is {(−5, −3), (1, −1), (2, 0), (3, 1)} The domain of the function is {−3, −1, 0, 1}; the range of the function is {−5, 1, 2, 3}. The domain of the inverse function is {−5, 1, 2, 3}; the range of the inverse function is {−3, −1, 0, 1}. ■ Ϫ4 Figure 51 Figure 51 shows the one-to-one function (in blue) and its inverse (in red). NOW WORK Domain of f Range of f f Figure 52 Remember, if f is a one-to-one function, it has an inverse f −1 . See Figure 52. Based on the results of
−1 are symmetric with respect to the line y = x, the inverse function f −1 can be obtained by interchanging the roles of x and y in f . If f is defined by the equation y = f (x) 36 Chapter P • Preparing for Calculus then f −1 is defined by the equation x = f (y) Interchange x and y The equation x = f (y) defines f −1 implicitly. If the implicit equation can be solved for y, we will have the explicit form of f −1 , that is, y = f −1 (x) Steps for Finding the Inverse of a One-to-One
(Ϫ π3 , Ϫ 1) Replace x by 3x; horizontal compression 1 by a factor of Ϫ. 3 (a) f (x) ϭ cos x π Ϫ3 53 Trigonometric Functions π 2 2π 3 x π 5π 6 ( π3 , Ϫ 1) (b) g(x) ϭ cos (3x) Figure 73 ■ From the graph, we notice that the period of g(x) = cos(3x) is NOW WORK 2π . 3 Problems 27 and 29. In general, if ω > 0, the functions f (x) = sin(ωx) and g(x) = cos(ωx) have 2π period T = . If ω > 1, the graphs of f (x) = sin(ωx) and g(x) = cos(ωx) are ω horizontally compressed and the period
27. tan θ = − 28. sec = −2 3 2 = Graphing technology recommended CAS = Computer Algebra System recommended 64 Chapter P • Preparing for Calculus 1 2 29. 2 sin θ + 3 = 2 30. 1 − cos θ = 31. sin(3θ ) = −1 32. cos(2θ ) = 33. 4 cos2 θ = 1 34. 2 sin2 θ − 1 = 0 35. 2 sin2 θ − 5 sin θ + 3 = 0 36. 2 cos2 θ − 7 cos θ − 4 = 0 37. 1 + sin θ = 2 cos2 θ 38. sec2 θ + tan θ = 0 39. sin θ + cos θ = 0 40. tan θ = cot θ 41. cos(2θ ) + 6 sin2 θ = 4 42. cos(2θ) = cos θ 43. sin(2θ ) + sin(4θ ) =
function f (x) = x 2 − 5x at: (a) c = 2 (b) any real number c Solution (a) For c = 2, f (x) = x 2 − 5x and f (2) = 22 − 5 · 2 = −6 The rate of change of f at c = 2 is f (2) = lim x→2 f (x) − f (2) (x 2 − 5x) − (−6) x 2 − 5x + 6 = lim = lim x→2 x→2 x −2 x −2 x −2 (x − 2)(x − 3) = lim (x − 3) = −1 x→2 x→2 x −2 = lim (b) If c is any real number, then f (c) = c2 − 5c, and the rate of change of f at c is f (c) = lim x→c f (x) − f (c) (x 2 − 5x) − (c2 − 5c) (x 2 − c2 ) − 5(x − c) = lim =
