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7: Suppose that (x) is defined for all in an open interval / with a el. (a) Prove the Decreasing Test: il l') <0 for all re I then / is decreasing in l. (4 16 (1) Prove Permat's Theorem: il l'(a) exists and then l'(a) = 0. has a local maximum or minimum at x =

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18 Jan 2018
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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) fΒ«+ (x) = x+2 ; (c) fΒΊ+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.

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.

[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) >

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