Math 1210
Quiz 3
January 24th, 2014
Answer the following three (3) questions. Every question is worth the same.
You may use scratch paper, but you can only turn in this sheet. Please
justify your answer in the space provided: writing the solution without any
explanation is not enough. No cell phones, calculators, notes, or music
players are allowed during the quiz.
Name:
UID:
√
1. Consider the function f (t) =
t−1
.
(t−1)(t+1)(t−2)
(i) What is the domain of f (i.e. where is f defined?)
(ii) At which points is the function discontinuous? Are the discontinuities removable /
jumps / vertical?
(iii) For every removable discontinuity, find the limit.
(iv) For every jump or vertical discontinuity, find the corresponding one-sided limits.
Solution:
(i) The domain of the function f is all of [0, ∞], except for the points
√ t = 1 and t = 2
where the denominator is 0. If t < 0, f (t) is not defined because t is not defined.
(ii) The function is discontinuous at the points t = 1 and t = 2 because the denominator
is zero. The denominator is also zero at t = −1, but this point is not in the domain
of f in the first place.
√
Evaluating at t = 2, wee see that f (2) =
at t = 2.
2−1
,
0
so there is a vertical discontinuity
Evaluating at t = 1, wee see that f (1) = 00 , so wee need to compute the limit. If
the limit exists, the discontinuity will be removable.
(iii) We evaluate the limit as t → 1:
√
√
t−1
t−1
√
= lim √
lim f (t) = lim
t→1 ( t + 1)( t − 1)(t + 1)(t − 2)
t→1
t→1 (t − 1)(t + 1)(t − 2)
1
1
= lim √
= √
t→1 ( t + 1)(t + 1)(t − 2)
( 1 + 1)(1 + 1)(1 − 2)
1
= −
4
(iv) We compute the left-hand and right-hand limits. As t → 2+ , every factor in the
definition of f (t) is positive, so limt→2+ f (t) = +∞.
As t → 2− , every factor in the definition of f (t) is positive except for t − 2 which
is negative, so limt→2− f (t) = −∞
2. State the intermediate value theorem and use it to show that the equation (cos t)t3 +
6 sin5 t − 3 = 0 has a solution between t = 0 and t = 2π (you don’t have to find the
solution, just show that there is one).
Solution: The intermediate value theorem states that if f is a continuous function which
is defined and is continuous over an interval [a, b], then for any number M ∈ [f (a), f (b)]
there exists some c ∈ [a, b] such that f (c) = M . In particular, if f (a) and f (b) have
opposite signs then M = 0 lies in between, so there exists some c ∈ [a, b] such that
f (c) = 0.
Consider the function f (t) = (cos t)t3 + 6 sin5 t − 3, so that a solution of the equation
(cos t)t3 + 6 sin5 t − 3 over [0, 2π] is a number c ∈ [0, 2π] such that f (c) = 0. The function
f is continuous everywhere, so in particular it is continuous over [0, 2π]. Moreover, we
have that
f (0) = (cos 0) · 03 + 6 sin5 0 − 3 = −3 < 0
f (2π) = (cos 2π) · (2π)3 + 6 · sin5 (2π) − 3 = 8π 3 − 3 > 0
Since f (0) < 0 < f (2π) and f is continuous over [0, 2π], the intermediate value theorem
guarantees that there exists c ∈ [0, 2π] such that f (c) = 0
3. Find the slope of the tangent line to the curve y =
1
x−1
at the point (0, −1).
Recall that if a curve is given by an equation y = f (x), the slope of the tangent line at
a point (c, f (c)) is given by the limit
f (c + h) − f (c)
h→0
h
lim
Solution: The slope of the tangent line at the point (0, −1) is given by the limit
f (x) − f (0)
= lim
lim
x→0
x→0
x−0
1
x−1
1
1
− 0−1
+1
x
1
x−1
= lim
= lim
= lim
= −1
x→0
x→0
x→0
x−0
x−0
x(x − 1)
x−1
The equation of the tangent line will be given by y = −x − 1
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