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Department
Mathematics
Course
MATA30H3
Professor
Sophie Chrysostomou
Semester
Winter

Description
University of Toronto at Scarborough Department of Computer & Mathematical Sciences MAT A30Y Solution to Assignment #4 A. Homework problems from the lectures: √ 1. (a) Let f(x) = sinx, g(x) = 1 − x . Find f ◦ g(x) and g ◦ f(x) and their domains. √ 2 solution : (f ◦ g)(x) = f(g(x)) = sin(g(x)) = sin( 1 − x ) with domain the set {x ∈ R|x ∈ domg and g(x) ∈ domf}. The domain of g is: domg = {x ∈ R|1 − x ≥ 0} 1 − x ≥ 0 ⇐⇒ x ≤ 1 ⇐⇒ |x| ≤ 1 ⇐⇒ −1 ≤ x ≤ 1 ∴ domg = [−1, 1] Since the domain of f(x) = sinx is R, then g(x) ∈ domf for all x ∈ domg. Thus, domf ◦ g = [−1, 1]. p p 2 √ (g ◦ f)(x) = g(f(x)) = 1 − f(x) = 1 − sin x = cos x = |cosx| with domain the set {x ∈ R|x ∈ domf and f(x) ∈ domg}. √ Since f(x) = sinx then domf = R. Since g(x) = 1 − x = [−1, 1] and −1 ≤ sinx ≤ 1 then f(x) ∈ domg for all x ∈ domf = R. Thus, domg ◦ f = R. x − 2x + 3 (b) If h(x) = 2 2 find f and g such that f(x) ▯= x and g(x) ▯= x and h = f ◦g. (x − 2x + 4) solution : x − 2x + 3 (x − 2x + 1) + 2 (x − 1) + 2 h(x) = 2 2 = 2 2= 2 2. (x − 2x + 4) ((x − 2x + 1) + 3) ((x − 1) + 3) 2 x + 2 Let g(x) = (x − 1) and f(x) = 2, then f ◦ g = h. (x + 3) 1 2. A) From the graph of cosx (in blue), we get the graph of secx (in green): 5 4 3 2 1 ▯π 0 π ▯1 ▯2 ▯3 ▯4 We need a one to one function so define f(x) = secx with domain [0,π/2) ∪ [π,3π/2) and range (−∞,−1] ∪ [1,∞). Its graph is in bold green color. Then f is one to one and y = f −1 (x) = sec −1 x = arcsecx for x ∈ (−∞,−1]∪[1,∞) and range [0,π/2)∪[π,3π/2). To get the graph of f −1 we get the mirror image of the graph of f at the mirror y = x: The graph of arcsec x π ▯5 ▯4 ▯3 ▯2 ▯1 0 1 2 3 4 B) Similarly From the graph of tanx (in blue), we get the graph of cotx (in green): 2 5 4 3 2 1 ▯π 0 π ▯1 ▯2 ▯3 ▯4 we define f(x) = cotx with domain (0,π) and range R. Then: y = f −1(x) = cot1 x = arccotx for (x ∈ R) and range (0,π). −1 To get the graph of f we get the mirror image of the graph of f at the mirror y = x: The graph of arccot x π ▯5 ▯4 ▯3 ▯2 ▯1 0 1 2 3 4 " ! # √ 3 3. Simplify sin arccos 2 + arctan(1) . solu"ion : √ ! # 3 π π sin arccos + arctan(1) = sin + 2 6 4 π π π π . = sin 6 cos 4 + cos6 sin 4 √ √ 1 1 3 1 1 + 3 = ▯√ + ▯√ = √ 2 2 2 2 2 2 3 Stewart. Section 1.6: 4 Stewart. Section 1.6: 5 Stewart. Section 1.6 & Appendix D: 6 C. Questions from topics in Calculus Review Manual (CRM): 1. Read pages 22-23. 2. Page 24: # 4, 14. (x − 5x + 4)(x + 2) (4.) Find all x satisfying ≥ 0 (x + 3)(2x + 1) 2 solution : Let F(x) = − 5x + 4)(x + 2) (x + 3)(2x + 1) A. & B. ) Factor and simplify: (x − 5x + 4)(x + 2(x − 4)(x − 1)(x + 2) F(x) = 2
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