Mathematics 1560 Chapter Notes - Chapter 2: The Sketch, Eval, Negative Number

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19 Jan 2018

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Additional practice: given f (x) = x, g(x) = x2, and h(x) = e2x, compute: (a) f (f (x)) =p x = (x1/2)1/2 = x1/4 = 4 x. (b) f (g(x)) = x2 = |x|. Note that it is only true that x2 = x when x 0. The function f (x) is the positive square root function. If x < 0 we have x2 = x. For example, if x = 2, then x2 = 4, and x2 = 4 = 2 = ( 2) = x. (c) g(f (x)) = ( x)2 = x. Since f ( x) = (b) g(x) = x|x| ( x)3 ( x)2 + 1. We have h( x) = cos(( x)5) = cos( x5) = cos(x5) = h(x), since cosine is an even function. Assigned problems: find the domain of the following functions.

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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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