MATH 181 Study Guide - Final Guide: Antiderivative, 4Dx

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turquoisegnu128

13 Dec 2018

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Problem 1 solution: the graph of a function g(x) is given below. Let g be an antiderivative for g on the interval [0, 5] with g(1) = 2. Solution: since g is an antiderivative of g we know that g (x) = g(x). G(x) = ! g(x) dx where g(x) = x 1 if 0 x < 3 if 3 x < 4. 2x + 10 if 4 x 5. 2: on the interval 0 x < 3, we have g(x) = x 1. The value of c1 is found by using the fact that g(1) = 2. G(1) = 2 (1)2 1 + c1 = 2. Therefore, on the interval 0 x < 3 we have g(x) = 1: on the interval 3 x < 4, we have g(x) = 2. The value of c2 is found by ensuring continuity of g(x) at x = 3. G(x) x 3 % 1 lim x2 x +

Related Questions

1. Expressthe solution set of the inequalityx^3-x^2+x-1>0<?xml:namespace prefix = v ns ="urn:schemas-microsoft-com:vml" /> <?xml:namespace prefix = ons = "urn:schemas-microsoft-com:office:office" /> ininterval notation.

2. Find the value ofk such that the line kx-3y=10 is perpendicular to the line2x+3y=6.

3. Find the point(s)of intersection of the graphs of y=x-1 and2x^2+3y^2=11.

4) For f(x)=2x^2-4x+1, find and simplifyf(a+h)-f(a)/h

5) For f(x)=x^2+x and g(x)=-4x+3 , find andsimplify:

a) (f*g)(x)

b)(g*f)(x)

c)(g*g)(x)

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.

MATH 181 Study Guide - Final Guide: Antiderivative, 4Dx

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