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[1] 6. (a) State the domain and range of tanh r. Solution: The domain is (-0,00) and the range is (-1,1). [1] (b) State the domain and range of tanh-r. Solution: The domain is (-1,1) and the range is (-0,0). [4] (c) Prove that tanh - 7 = in (1+). Solution: If y = tanh-1, then I = tanh y = ee. Thus, el – EU elle- (e" + e-").r = e-e-Y rey - y = -e-yr-e- e24(1 - 1) = -r-1 220 = 7-1 2y = ln (1+) So, tanh- = (1+2)

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Trinidad Tremblay
Trinidad TremblayLv2
5 Aug 2018
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[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

4. Sketch the graphs of the following functions in (a) through (c). You should explicitly address each of the following aspects in your solution. (1) Is the function even, odd, or neither under reflection about the y axis? (ii) What are the domain and range of the function? (iii) Does the function have any asymptotes (vertical, horizontal, or slant)? (iv) What is the y-intercept? (v) Find all roots (a-intercepts) of the function. If the function has no roots, state this explicitly, along with your reasoning. (vi) Find any global (absolute) extrema (maxima or minima). What is the maximal or minimal value? If there are no global extrema, state so explicitly, along with your reasoning (b) r(t) = 24 – 16

Problem 4 Circle your answer. a) (6 points) If f(x) = 2x + 3x2 - 361 + 5, then the tangent line to the graph y= f(x) is horizontal when i) 1 = -2,3 ii) I = 2,-3 iii) r = 0 iv) not listed b) [6 points] Suppose Given h(x) = f(g(T)), where g(2) = 3, g'(2) = 7, f'(2) = -1 and 1'(3) = -2. Then h'(2) is equal to i) - 1 ii) -2 iii) - 14 iv) not listed (x+3)(1 - 1)3 c) (6 points) The function f(1) == (x2 - 2x + 1)(x + 3) wa has vertical asymptotes at i) x = -3,0 ii) = -3,0,1 iii) x = 0,1 iv) not listed d) (6 points] The solution to the initial value problem v'(x) = 473 – cos(x) +e", y(0) = 3 is i) y(x) = x + sin(x) + e* + 2 ii) y(I) = x4 – sin(x) +et+2 iii) y(x) = 2* – sin(1) + et + 3 iv) not listed

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