QUIZ ANSWERS

BRIEF QUIZ ANSWERS FOR CALCULUS I STUDENTS

In this document you will find brief answers to quiz questions. If the quiz question is a homework problem, and the answer is in the back of the book, then the brief answer that you will see below will be only ``See the back of the book.''

QUIZ

Week 1 (Aug 25)
Week 2 (Sep 1)
Week 3 (Sep 8)
Week 4 (Sep 15)
Week 5 (Sep 22)
Week 6 (Sep 29)
Week 7 (Oct 6)
Week 8 (Oct 13)
Week 9 (Oct 20)
Week 10 (Oct 27)
Week 11 (Nov 3)
Week 12 (Nov 10)
Week 13 (Nov 17)
Week 14 (Nov 24)
Week 15 (Dec 1)

Week 1

No quiz.

Week 2

See the back of the book.
(1/2)tan((5Pi/6)-(x/2)) + 3.
See the back of the book.
See the back of the book.

Week 3

See the back of the book.
domain = (-2,2). range = (-infinity,ln(4)].
See the back of the book.
One property that distinguishes e from other real numbers is the following: the tangent line to the function y = log_a(x) at x = 1 has slope = 1 if and only if a = e. Similarly, the tangent line to the function y = a^x at x = 0 has slope 1 if and only if a = e.

Week 4

See the back of the book.
See the back of the book.
See the back of the book.
For any epsilon greater than zero, there is a delta greater than zero such that: if x is not a and |x-a| is less than delta, then |f(x)-L| is less than epsilon.
There ARE functions that satisfy the false definition of limit. For example, if you choose f(x) = x^2, L = -2, and delta = 1, then the definition is satisfied. (Roughly the false definition says: f(x) is never ``near'' the value L unless it is exactly equal to L. ``near'' L means: within delta units of L.)

Week 5

Test 1.

Week 6

See the back of the book.
See the back of the book.
f'(a) = lim_{x goes to a} (1/x^2 - 1/a^2)/(x-a) = lim_{x goes to a} (a^2-x^2)/(x^2a^2(x-a)) = lim_{x goes to a} -(a+x)/(x^2a^2) = -2a/a^4 = -2/a^3.
Differentiate:

See the back of the book.
See the back of the book.
See the back of the book.

Week 7

If the line passes through (0,-4) and is tangent at (a,f(a)), then the slope of the line between these points, which is (f(a)-(-4))/(a-0), should equal f'(a). Thus, we need to find a for which f'(a) = (f(a)+4)/a. Since f(x) = x^3 we have f'(x) = 3x^2. This shows that we need a value of a for which 3a^2 = (a^3+4)/a. Solving yields a = 2^(1/3). Thus, the line is y = f'(a)(x-a) + f(a) = 34^(1/3)x -4.
See the back of the book.
See the back of the book.
Use induction.

Basis step (n=0). (1+x)^0 = 1 = 1 + 0x.
Inductive step. Assume that (1+x)^k is greater or equal to 1 + kx. Then (1+x)^(k+1) = (1+x)(1+x)^k, which is greater or equal to (1+x)(1+kx) = 1 + (k+1)x + kx^2, which is greater or equal to 1 + (k+1)x.

Week 8

See the back of the book.
See the back of the book.
dy/dx = cos(x)(1-2sin(x)), so when x is between zero and 2Pi the derivative is zero only at x = Pi/6, Pi/2, 5Pi/6 and 3Pi/2.
When x = 2, then 5^y = 3^2 + 4^2 = 25, so y = 2. At (x,y) = (2,2) the derivative is ln(3^(3^2)4^(4^2))/ln(5^(5^2)).

Week 9

See back of book.
See back of book.
f'(x) = (x^(x^x))(x^x)((1/x) + ln(x) + (ln(x))^2), so f'(e) = (e^(e^e))(e^e)((1/e) + 2).
Variables: y = distance between tips of clock hands, a = angle between hands, t = time. Want: dy/dt at a = Pi/6. The relationship between the variables is given by the Law of Cosines: y^2 = 8^2 + 4^2 - 284cos(a). Need to find dy/dt: 2y(dy/dt) = 284sin(a)(da/dt). To use this equation to solve for dy/dt, we first need to find y and da/dt when a = Pi/6. The value of y at Pi/6 is 4sqrt(5-2sqrt(3)); this can be found with the Law of Cosines. The value of da/dt is -(11/12)(2Pi/60) rad/min at any time of the day, since there are 60 minutes per hour and in one hour the minute hand advances 11/12 of the way around the circle compared to the hour hand.

Week 10
Test 2.

Week 11

critical point at x=1 (absolute max); inflection point at x=2; CD on (-infinity,2), CU on (2,infinity).
Sketched in class. (Check your notes from November 5.)
p(x) = -(x-1)^4 has an absolute max at x = 1 (since p(1)=0 and p(x) is negative whenever x is not 1), but the second derivative test fails to classify this point as a local max since p''(0) = 0.
g'(x) = 2abs(x), which is not differentiable at x=0. Thus g''(0) does not exist. However g''(x) is positive when x is positive, and g''(x) is negative when x is negative, so the concavity of the graph of y = g(x) changes at x=0. This shows that g has an inflection point at x=0.

Week 12

See the back of the book.
L'Hospital's Rule implies that a function that has a jump discontinuity cannot be the derivative of any function. (Check your notes from November 5.)
L(x) = (1/2)x - (5/4).
y = (-5/3)x + 10.

Week 13

L = 4*(1+2^(2/3))^(3/2).
See the back of the book.
False. The MVT implies that there is some c between 1 and 3 such that f'(c) = (f(3)-f(1))/(3-1) = 1. Thus we cannot have f'(x) greater than 1 for all x.
lim_{n goes to infinity} (1^3 + 2^3 + ... + n^3)/n^4.

Week 14

Thanksgiving.

Week 15

See the book.
See the back of the book.
See the back of the book.
See the back of the book.

Last modified on Aug 30, 1999.