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Mathematics and Statistics

MATH 1505

William Pietro

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.
′

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