# Instructor's Solution Manuals to Calculus: Early by Robert T. Smith and Roland B. Minton

By Robert T. Smith and Roland B. Minton

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Additional resources for Instructor's Solution Manuals to Calculus: Early Transcendental Functions

Sample text

16. e4x−2 = f (g(x)) when f (x) = ex and g(x) = 4x − 2, for example. 3 = f (g(h(x)) when f (x) = 3/x, sin x +√2 g(x) = x, and h(x) = sin x + 2, for example. √ 18. e4x + 1 = f (g(h(x))) when f (x) = x, g(x) = x + 1, and h(x) = e4x , for example. 17. √ 19. cos3 (4x − 2) = f (g(h(x))) when f (x) = x3 , g(x) = cos x, and h(x) = 4x − 2, for example. 20. ln x2 √ + 1 = f (g(h(x))) when f (x) = ln x, g(x) = x, and h(x) = x2 + 1, for example. 6. TRANSFORMATIONS OF FUNCTIONS 31 2 21. 4ex − 5 = f (g(h(x))) when f (x) = 4x − 5, g(x) = ex , and h(x) = x2 , for example.

One possibility: -2 -3 27. Numerical and graphical evidence show that 48 CHAPTER 1. LIMITS AND CONTINUITY x2 + 1 x+1 and lim 2 do not x→1 x − 1 x→2 x − 4 exist (both have vertical asymptotes). Our conjecture is that if g(a) = 0 and f (a) = 0, f (x) does not exist. lim x→a g(x) the limits lim 31. 001 30. From the values shown in table below, we can conclude f (x) tends to infinity as x tends to 0. 7182818 x→0 x+1 sin x 28. lim 2 = 0 and lim = 0. If the x→π x x→−1 x + 1 numerator f (a) = 0, and the denominator f (x) g(a) = 0, then the limit lim = 0.

The result can be checked, and a graphing calculator can find them by graphing y = x3 and y = sin x on the same axes and finding the intersection points. 1 y 0 -1 0 1 2 3 x -1 -2 38. The graph shows two zeros. Squaring both sides gives x2 +1 = x4 −2x√2 +1, or 0 = x4 −3x2 . The solutions are x = ± 3. ) -3 28. No intercepts, extrema, or asymptotes. Function only defined for x > 0. 4 2 39. Let h be the height of the telephone pole. 7 feet. 40. The triangle in the first quadrant with adjacent side 1 and hypotenuse 5 has opposite side √ √ 24 24, so sin θ = 5 .