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MAT 125 Study Guide - Final Guide: Intermediate Value Theorem, Asymptote

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turquoisegnu128
28 Sep 2018
School
Stony Brook University
Department
Mathematics
Course
MAT 125
Professor
All

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MAT 125 Full Course Notes
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MAT 125 Full Course Notes
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Document Summary

Spring 2012: in parts (a) (c), compute the given limits. (a) lim x 1 x2 x 2 x2 1. 2 (x + 1)(x 2) (x + 1)(x 1) lim x 1 x 2 x 1. 3x + 1 (b) lim x x2 4x + 5. Because the degree of the denominator is greater than the degree of the numerator, the limit is zero. 9 + h 3 (cid:19)(cid:18) lim h 0 h (cid:19) 6 (d) the limit in part (c) equals f(cid:48)(a) for: a = 9 , f (x) = (this is not the only possible answer: one could also choose, for instance, a = 0, f (x) = 9 + x. ) x: suppose lim x 2 f (x) = 2 and lim x 2 g(x) = 3. This follows directly from the properties of limits. Answer: 4 (cid:0)f (x)(cid:1)2g(x) (cid:2)f (2x) 2g(2x)(cid:3) (a) lim x 2 (b) lim x 1 (c) lim x 2.

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Related Questions

1. (20 pts) The graph of a function f is given in the figure below. Use the graph of f to answer the questions below. y = f(x) -64 - - - - - - - - - - - - (I) (1 pt) Pind the domain of f. Domain of f = (-0, 3) U (3, +co) (II) (1 pt) Find the range of f. Range of f = (-, +O) (a) (1 pt) lim f (x) = 0 (b) (1 pt) f(3) = DNE (does not exist) (c) (1 pt) f (5) = 2 (a) (1 pt) lim f (x) = 0 (0) (1 pt) lim f (x) = -- (f) (1 pt) lim f (x) = DNE 0 1 (g) (1 pt) lim f (x) = + (h) (1 pt) lim f (x) = -6 X-++00 (IV) (2 pts) Find all vertical asymptotes. (V) (2 pts) Find all horizontal asymptotes. (VI) (6 pts) Determine the intervals of continuity for f. (-0, 0], (0, 3), (3, 5), (5, +00)

1. (25 pts) The graph of a function f is given in the figure below. The domain of fis (-, +.). Use the graph of fto answer the questions below. yax) y=f(x) -3 - 2 3 4 (I) (1 pt) Find the range of f. (II) Find the following values. (Note: Possible answers include + , -, or "does not exist".) (a) (1 pt) (x) = 5 (b) (1 pt) lim f (x) = 5 (0) (1 pt) Tim f (x) = 5 (d) (1 pt) f(3) = 2 (e) (2 pts) lim f (x) = + 0 x-2- (8) (2 pts) 11m = (x) = 0 (g) (2 pts) lim f (x) = DOES NOT EXIST x+-2 (h) (2 pts) Lim f (x) = 0 (i) (2 pts) lim f (x) = + X +00 (III) (2 pts) Find all vertical asymptotes. x = -2 (IV) (2 pts) Find all horizontal asymptotes. y = 0 (v) (2 pts) Find all slant asymptotes. | y = x + 2 (VI) (4 pts) Determine the intervals of continuity for f. (-00, -2), [-2, 3), and (3, +c)

4. (21 pts) (I) (6 pts) Determine the value of a constant b for which the function g is continuous at 0. Show your work. Justify your answer! 2 x +b if x<0, X-5 g(x) = if x 20, x#4. We have to show that lim g (x) = g (0). g(0) = 0 +18 = -1; lim X0* 0 + 16 ) = lim x + 16 **0* x2 - 160 - 16- Since lim g (x) has to exist (in order for g to be continuous at x = 0), it has to be true that x-0- x0+ X- 0 lim g(x) - lim g(x) = lim g(x). Therefore, -1, and , so b = 5. so, if b = 5, lim g (x) = g(0), and therefore, g is continuous at x = 0. (II) (4 pts) Evaluate the limit. Show your work! 2. b limx- (a) lim g(x) = lim *+- 2 x + b X - 5 : 1 * +16 10 g (x) = lim x++ - Elim x2 - 16 x++ (x - 10). OR 2x+5. 2; (a) lim g (x) = lim x--00 X- (b) lim x++ 2 + 16 g(x) = lim **** x2 - 16 X-+ (III) (4 pts) Determine whether g has any horizontal asymptotes. If so, write the equation (or equations) of all horizontal asymptotes. y = 2, by part (II) (a) y = 0, by part (II) (b) (IV) (7 pts) Determine whether g has any vertical asymptotes. If so, write the equation (or equations) of all vertical asymptotes. Show your work. Justify your answer! 5 + 16 There is no vertical asymptote at x = 5, because lim g(x) = x+5 52-16' -8 + b (x) = there is no vertical asymptote at x = -4, because lim X -4 - 4 - 5 But, lim g(x) = lim + 16 x+4+ x2 - 16 lim_ (x + 16) -20 x+4+ (x - 4) 60 (x + 4) +8 (or the numerator goes to 20, and the denominator is positive and goes to o, as x + 4*) = +00 So, the function g has only one vertical asymptote, x = 4.