MATH 171 Midterm: Math 171 TAMU Fall 12 Exam2ExtraPointsSolutions

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turquoisegnu128

31 Jan 2019

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Nov. 21, 2012: (5) prove, using epsilons and deltas, that if lim x 4 f (x) = 6, and lim x 4 g(x) = 2, then lim x 4. Then there are numbers 1 > 0 and 2 > 0 such that if. 0 < |x 4| < 1 then |f (x) 6| < Then, > 0, and if 0 < |x 4| < , |f (x) + g(x) 8| = |[f (x) 6] + [g(x) 2]| |f (x) 6| + |g(x) 2| . 2: (5) show for all x, y that |cos(x) cos(y)| |x y|. This is a straight forward application of the mean value theorem, and the fact that | sin x| 1 for all x. Find the rate of change of the length of an edge of the cube when the volume is 27 cubic meters. The volume of a cube is equal to the cube of its edge length.

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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.

[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) >

MATH 171 Midterm: Math 171 TAMU Fall 12 Exam2ExtraPointsSolutions

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