7 Questions with Solution of Midterm Exam 1 - Calculus | MATH 1A, Exams of Calculus

Download 7 Questions with Solution of Midterm Exam 1 - Calculus | MATH 1A and more Exams Calculus in PDF only on Docsity! Midterm 1, Math 1A, section 1 solutions 1. Let F (x) = √ 2 + x, G(x) = √ 2− x. Find F − G, FG, F/G, and G ◦ F , and find their domains. Determine which of these functions is even, odd, or neither. Solution: First, we find the domain of F and G. If F (x) = √ 2 + x, then we must have 2 + x ≥ 0, so x ≥ −2. If G(x) = √ 2− x, then 2− x ≥ 0, so x ≤ 2. We have F (x)−G(x) = √ 2 + x− √ 2− x. The domain is −2 ≤ x ≤ 2. FG(x) = √ 2 + x √ 2− x = √ (2 + x)(2− x) = √ 4− x2. Then FG has domain −2 ≤ x ≤ 2, and FG is even since FG(−x) = √ 4− (−x)2 = √ 4− x2 = FG(x). (F/G)(x) = √ 2+x√ 2−x . This has domain −2 ≤ x < 2, since the denominator cannot = 0. F/G(x) is neither even nor odd since its domain is not symmetric about x = 0. G ◦ F (x) = √ 2− √ 2 + x. For G ◦ F to be defined, we must have F (x) lies in the domain of G(x), so F (x) ≤ 2. Then √ 2 + x ≤ 2, so we have 0 ≤ 2 + x ≤ 4, and thus −2 ≤ x ≤ 2. G ◦ F is neither odd nor even, since since G ◦ F (2) = √ 2− √ 2 + 2 = 0, while G ◦ F (−2) =√ 2− √ 2− 2 = √ 2, so 0 = G ◦ F (2) 6= ±G ◦ F (−2) = ± √ 2. 2. Draw the graph of y = x2. Use the graph to find a number δ such that if |x − 1| < δ, then |x2 − 1| < .96 = 2425 . Label the corresponding intervals on your graph. Solution: The inequality |x2− 1| < 2425 is equivalent to − 24 25 < x 2− 1 < 2425 . Adding 1 to each part of the inequality, we obtain 125 = 1− 1 25 < x 2 < 1 + 2425 = 49 25 . Since the positive square root preserves inequalities, this is equivalent to √ 1 25 = 1 5 < x < 7 5 = √ 49 25 for x > 0. Now, we subtract 1 from both sides, obtaining −45 = 1 5 −1 < x−1 < 2 5 = 7 5 −1. Thus, we see that if we let δ < 2 5 , then if |x − 1| < δ, we have −45 < − 2 5 < x − 1 < 2 5 , and therefore from the above reversible derivations, we get |x2 − 1| < 2425 . 3. Let f(x) = x+8 x2−4 (a) What is the domain of f? (b) Find f(1), f(−3), and the x- and y-intercepts of f . (c) Is f even, odd, or neither? Give an explanation. (d) Find lim x→∞ f(x), lim x→2+ f(x), lim x→2− f(x). What are the asymptotes of y = f(x)? (e) Sketch all of the points and asymptotes you have found from the previous parts on a graph. Then sketch the graph of y = f(x) on the same graph. Solution: (a) f(x) is defined when the denominator is non-zero, so when x2 − 4 = (x− 2)(x+ 2) 6= 0, which is equivalent to x− 2 6= 0 and x+ 2 6= 0. Thus, the domain of f(x) is x 6= ±2. (b) f(1) = 1+8 12−4 = 9 −3 = −3. f(−3) = −3+8 (−3)2−4 = 5 5 = 1. f(0) = 8 −4 = −2. The y-intercept is obtained by setting f(x) = 0, which happens when the numerator is zero, and therefore x = −8. (c) f is neither even nor odd, since the denominator is even, but the numerator is neither odd nor even. Alternatively, one may use that f(0) = −2 6= 0, so f is not odd, and f(8) = 415 > 0 = f(−8), so f is not even. (d) lim x→∞ f(x) = lim x→∞ x+8 x2−4 = limx→∞ 1/x+8(1/x)2 1−4(1/x)2 = 0 1 = 0, where we are using the fact that lim x→∞ 1/x = 0, and we may plug this into a limit of a continuous function. When −2 < x < 2, we have x2 − 4 < 0, x+ 8 > 6, and lim x→2− x2 − 4 = 0. Thus, f(x) < 0. So we have lim x→2− x+8 x2−4 = −∞. 2