Study Guides (248,269)
Canada (121,449)
York University (10,192)
MATH 1505 (58)
Midterm

Test 2 2014

7 Pages
395 Views
Unlock Document

Department
Mathematics and Statistics
Course
MATH 1505
Professor
William Pietro
Semester
Winter

Description
YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH 1505 6.00 E Test #2 February 28, 2014 Solutions Family Name (print): Given Name: Student No: Signature: INSTRUCTIONS: 1. Please write your name, student number and final answers in ink. 2. This is a closed-book test, duration – 50 minutes. 3. Non-programmable, non-graphing calculators are permitteNO INTERNET CON- NECTED DEVICES OR OTHER AIDS ARE PERMITTED. 4. There are six questions on seven pages. Fill in answers in designated spaces. Your work must justify the answer you give. Show your work on the space provided. If you need more space, use the back of a page and clearly indicate this fact on an original page each time when you use the back of a page for your work. 5. Remain seated until we collect all the test papers. 6. Please note that there are 60 points in the test. However, your test grade will be counted out of 50. Do the easiest questions first, GOOD LUCK! Question Points Scored 1 8 2 14 3 6 4 9 5 9 6 14 Total: 60 1. (4 + 4 pts) 1 (a) Determine whether the function f(x) = satisfies the conditions of the Mean-Value 1 + x Theorem on the interval [0;1]? If so, find the value of c that satisfies the conclusion of the Mean-Value Theorem. ANSWER: Yes. ′ d 1 1 f(x) is continuous on [0;1]; differentiable on (0;1) and f (dx 1 + x ) = − (1 + x) : So, by the Mean-Value Theorem there must exist c ∈ (0;1); such that f(1) − f(0) ′ 1 − 0 = f (c): Hence, we have 2− 1 1 1 1 2 = − 2 =⇒ − = − 2 =⇒ (1 + c) = 2: 1 − 0 (c + 1) 2 (c + 1) √ √ √ That is, c = ± 2 − 1: But − 2 − 1 < 0: Therefore, c = 2 − 1: x x (b) Determine whether the function g(x) = sin( ) + cos( ) satisfies the conditions of Rolle’s 2 2 Theorem on the interval [− ;3▯ ]? If so, find the value of c that satisfies the conclusion 2 2 of the Rolle’s Theorem. ANSWER: Yes. x x ▯ 3▯ ▯ 3▯ Indeed, g(x) = sin( ) + cos( ) is continuous on [− ; ]; differentiable on (− ; ) 2 2 2 2 2 2 and g(−▯ ) = sin(− ) + cos(− ) = − 1 + 1 = 0; 2 2 2 2 2 g(3▯ ) = sin(3▯) + cos(3▯) = 1 − 1 = 0: 2 2 2 2 2 Hence, by Rolle’s theorem there must exist c ∈ [− ; ] such that f (c) = 0: 2 2 Hence, we have ′ 1 c 1 c c c f (c) = sin( ) − cos( ) = 0 =⇒ sin( ) = cos( ) 2 2 2 2 2 2 c ▯ ▯ =⇒ = + 2▯n ∀n ∈ Z =⇒ c = + 4▯n ∀n ∈ Z: 2 4 2 ▯ Taking n = 0; we obtain c = : 2 2 2. (4 + 4 + 4 + 2 pts) Let y = f(x) = 2 + x − x : The first and second derivatives are (x − 1) x − 5 14 − 2x f (x) = 3 and f (x) = 4: (x − 1) (x − 1) 1 (a) Determine the domain, x and y-intercepts, horizontal and vertical asymptotes of y = f(x) (if any). ANSWER: 2 Dom(f(x)) = {x ∈ R | (x − 1) ̸= 0} = {x ∈ R | x ̸= 1} = (−∞;1) ∪ (1;∞): f(x) = 0 implies that x − x − 2 = 0 that is (x − 2)(x + 1) = 0: So, x = −1 and x = 2 are x-intercepts. f(0) = 2; so, the function has y-intercept at y = 2: lim f(x) = −1 − and lim f(x) = −1 : So, y = −1 is a horizontal asymptote. x→∞ x→−∞ lim f(x) = lim f(x) = ∞: So, x = 1 is a vertical asymptote. x→1 ▯ x→1 + (b) Determine the intervals where f(x) is increasing and where it is decreasing. Find the coordinates of the local extremum point(s) of f(x) (if any). ANSWER: f (x) does not exist at x = 1; however f(x) is not defined there. f (x) = 0 implies 9 9 x − 5 = 0; that is x = 5: And f(5) = − : So, (5;− ) is the only critical point. ′ 8 8 ′ f (x) > 0 on the intervals (−∞;1) and (5;∞); and f (x) < 0 on the interval (1;5): Hence, f(x) is increasing on the intervals (−∞;1) and (5;∞); and decreasing on the 9 interval (1;5): Therefore, the point (5;− ) is a local minimum point. 8 (c) Determine the intervals where the graph of f(x) is concave up and where it is concave down. Find the coordinates of inflection point(s) of f(x) (if any). ANSWER: f (x) does not exist at x = 1; however f(x) is not defined there. f (x) = 0 implies 10 10 2(7−x) = 0; that is x = 7: Since f(7) = − ; the point (7;− ) is a possible inflection 9 9 point. f (x) > 0 on the intervals (−∞;1) and (1;7); and f (x) < 0 on the interval (7;∞): Hence, f(x) is concave up on the intervals (−∞;1) and (1;7); and concave down on the intervals (7;∞): 10 Therefore, the point (7;− ) is an inflection point. 9 (d) Use all the obtained information to sketch the graph of y = f(x). What is the range of f(x)? ANSWER: From parts (a) and (b) it follows that the function has no local or global maximum, 9 and the local minimum point (5;− ) is also the global minimum point of the function. 8 Therefore, the range of f(x) is [− ;∞): 8 The scanned copy of the graph of y = f(x) is given on separate page. 3. (6 pts) A toxin is introduced into a bacterial colony, and t hours later, the population of the colony is given by −0:1t N(t) = 10000(8 + t)e : 2 Determine after how many hours is the population maximized. What is the maximum pop- ulation? Justify your answers. ANSWER: N(t) is continuous and we need to maximize N(t) on [0;∞): ′ −0:1t −0:1t −0:1t N (t) = 10000[e + (8 + t)e (−0:1)] = 1000e (2 − t): Hence, ′ N (t) = 0 =⇒ 2 − t = 0 =⇒ t = 2: Thus, t = 2 is a critical point. Since C(t) is differentiable, there is no other critical points. ′
More Less

Related notes for MATH 1505

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit