1 (10) 2 (10) 3 (10) 4 (10) 5 (10) 6 (10) Total (60)

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1 First Name: OSU Number: Last Name: Signature: OKLAHOMA STATE UNIVERSITY Department of Matematics MATH 2144 (Calculus I) Instructor: Dr. Matias Sculze MIDTERM 1 September 17, 2008 Duration: 50 minutes No aids allowed. Tis examination paper consists of 7 pages and 6 questions. Please bring any discrepancy to te attention of an invigilator. Te number in brackets at te start of eac question is te number of points te question is wort. Answer 5 of 6 questions. To obtain credit, you must give arguments to support your answers. For graders use: 1 (10) 2 (10) 3 (10) 4 (10) 5 (10) 6 (10) Total (60) Score Page 1 of 7

2 page 2 1. [10] True or False? Write a T (for true) or an F (for false) for eac statement. ( (a) lim 2x x 4 ) 8 x 4 x 4 = 2x limx 4 lim 8 x 4 x 4 x 4 (b) If p is a polynomial, ten lim x 1 p(x) = p(1). (c) If lim x a [f(x)g(x)] exists, ten it must be equal to f(a)g(a). (d) If lim x a f(x) = and lim x a g(x) =, ten lim x a [f(x) + g(x)] = 0. (e) If x = 1 is a vertical asymptote of y = f(x) ten f is not defined at 1. (f) If f is continuous at a, ten f is differentiable at a. (g) If f(x) > 1 for all x > 0 and lim x 0 + f(x) exists, ten lim x 0 + f(x) > 1. () If f (r) exists, ten lim x r f(x) = f(r). (i) Te equation x 10 10x = 0 as a root in te interval (0, 2). (j) A rational function can ave two different orizontal asymptotes. (a) F (limit law does not apply to infinite limits) (b) T (polynomials are continuous) (c) F (example: f(x) = x, g(x) = x 1, a = 0) (d) F (example: f(x) = x 2, g(x) = x 4, a = 0) (e) F (limits in te definition of te vertical asymptote ignore f(1)) (f) F (example: f(x) = x ) (g) F (example: f(x) = x + 1) () T (differentiable implies continous) (i) T (apply Intermediate Value Teorem to [0, 1])

3 page 3 2. [10] Give a simple example or wite N/A if tere is no suc example. (a) A polynomial tat is a power function. (b) A polynomial tat is not a rational function. (c) A rational function tat is not a polynomial. (d) An inverse trigonometric function tat is not an algebraic function. (e) A continuous function witout orizontal and vertical asymptotes. (f) A function wit 3 vertical asymptotes. (g) A root function wose domain does not include 1. () An algebraic function wit domain ( 1, 1). (i) A function wit infinitely many discontinuities. (j) A continuous but not differentiable function. (a) f(x) = 1 (= x 0 ) (b) N/A (by definition any polynomial is a rational function) (c) f(x) = 1 x (d) sin 1 (essentially any inverse trigonometric function works) (e) f(x) = x (f) f(x) = 1 (= 1 ) x 3 x (x ( 1))(x 0)(x 1) (g) f(x) = x () f(x) = 1/ 1 x 2 (i) f(x) = [x] (discontinuous at all integers) (j) f(x) = x

4 page 4 3. [10] (a) For f(x) = 2x2 18, find all asymtotes and te limits tat describe te asymptotic x 2 +2x 3 beavior of te function. (b) Find te orizontal asymptotes of te function f(x) = x 6 1 x 3 +7x 2 +4x 8. (a) Dropping te terms wit not igest exponents in te numerator and denominator of f (as explained in te lecture) yields y = 2x2 = 2 as orizontal x 2 asymptote for bot x and x. So we ave lim x ± f(x) = 2. To find te vertical asymptotes, we factorize and cancel factors if possibe: f(x) = 2x2 18 x 2 + 2x 3 + 3)(x 3) = 2(x (x + 3)(x 1) = 2x 3 x 1 So, x = 1 is te only vertical asymptote. As x 3 < 0 for x close to 1, we ave lim x 1 = and lim x 1 + =. (b) Dropping te terms wit not igest exponents under te root and in te denominator of f (as above) yields x 6 = x 3 = x /x wic as te same x 3 x 3 asymptotic beavior as f(x) for large x. So lim x f(x) x x /x = 1 and similarly lim x f(x) = 1. In oter words, y = 1 and y = 1 are two (different) orizontal asymptotes. Page 4 of 7

5 page 5 4. [10] (a) Find all values for a and b suc tat te function x 2 9 if x < 1 x 3 f(x) = (x a) 2 if 1 x < 2 2ax b if 2 x becomes continuous. (b) Is f differentiable for some coice of a and b? (a) First, note tat f(x) = x + 3 for x < 1. Continuity is clear at x 1, 2. Te following two conditions are equivalent to continuity at 1 and 2 respectively: 4 f(x) = f(1) = (1 a)2, x 1 (2 a) 2 f(x) = f(2) = 4a b. x 2 Te first equality gives a = 1 2, so a = 1 or a = 3. Ten te second equality reads 5 ± 4 = (1 ± 2) 2 = 4 8 b wic gives b = So eiter a = 1 and b = 13 or a = 3 and b = 11. (b) Note tat if f is continuous, ten f(1) = 4 by te first part. For f (1) = to exist, te corresponding left- and rigt-sided limits lim 0 f(1+) f(1) f(1 + ) f(1) lim 0 f(1 + ) f(1) lim = 1, 0 + (1 + a) 2 4 must be equal. But, for a = 1 2, te rigt-sided limit equals ( ± 2) 2 4 lim ± 4 = ±4 wic is not equal to te left-sided limit. So te answer is no. Page 5 of 7

6 page 6 5. [10] (a) Compute te derivative of f(x) = 1 x 1+x (b) Find te domains of f(x) and f (x). (a) f f(x + ) f(x) (x) x 1 x 1+x+ 1+x (using te limit definition). 0 (1 x )(1 + x) (1 x)(1 + x + ) (1 + x + )(1 + x) 1 x + x x 2 x 1 x + x + x 2 + x 0 (1 + x + )(1 + x) 2 0 (1 + x + )(1 + x) = 2 (1 + x) 2 (b) Bot domains are obviously R \ { 1}.

7 page 7 6. [10] Compute te limits. (a) lim x 4x 6 x x 3 +9 (b) lim t t 2 t 2t 2 +t+7 ( (c) lim x 0 x 4 cos x) 2 (Hint: use te Squeeze Teorem) ( ) (d) lim x π arctan 64x 2 π x 8π (e) lim x π sin(x + sin(x + sin(x + sin(x + sin x)))) (a) Note tat for x < 0, we ave x 3 = x 6. Using tis, we compute lim x 4x 6 x x 3 +9 x 4x 6 x q x x (x 3 +9)x 3 x 5 = x 3 t (b) lim 2 t t 2t 2 +t+7 t t2 = 1 2t 2 2 (c) We ave x 4 x 4 cos 2 x x4 and ( lim x 0 x 4 = 0. So, by te Squeeze Teorem, it follows tat also lim x 0 x 4 cos x) 2 = 0. ( ) ( ) (d) By Teorem 8 in Section 2.5, lim x π arctan 64x 2 π 2 64x = arctan lim 2 π x 8π x π. 8 64x 8π 64x But lim 2 π 2 x π 8 64x 8π x π (x+ π) = π and ence te result is arctan π = (e) By continuity, lim x π sin(x+sin(x+sin(x+sin(x+sin x)))) = sin(π +sin(π + sin(π + sin(π + sin π)))) = 0. End of examination Total pages: 7 Total marks: 60