Saint Marys University Introductory Calculus First Derivative Test Worksheet

Indeterminate Forms and L’Hôpital’s Rule
Required Reading: Section 4.5
Expressions such as 00 and ± ∞
∞ look like ordinary numbers but they’re not !
They are called indeterminate forms; they have the form of an ordinary number
but we can not assign them a value that is consistent with the usual rules for
the algebra of numbers.
These indeterminate forms can arise when we are trying to evaluate the
limit of certain functions, and in this section, we will learn how to deal with
indeterminate forms when they show up.
Example 1 The limit
lim
x→0
approaches the indeterminate form
3x − sin(x)
x
0
0
as x approaches 0.
What makes this example stand out? We have seen a denominator of zero
before while taking limits . . . why make a big deal of this now?
What makes the above example stand out from what we have seen in the
past is the following: you can not perform any sort of direct simplification to
evaluate the limit (like factoring something out of the numerator and denominator or multiplying numerator and denominator by a conjugate).
Okay, we can’t simplify directly first and evaluating the limit leads to an
indeterminate form, so limits of this variety don’t exist? That’s not necessarily
the case ! L’Hôpital’s Rule gives us a tool that can help us evaluate limits that
approach indeterminate forms:
Theorem 1 (L’Hôpital’s Rule) Suppose that f (a) = g(a) = 0, that f and g
are differentiable on an open interval I containing a, and that g 0 (x) 6= 0 on I if
x 6= a. Then
f (x)
f 0 (x)
lim
= lim 0
,
x→a g(x)
x→a g (x)
assuming that the limit on the right side of this equation exists.
A few quick notes about this theorem:
1
(i) This theorem also works if both f (a) → ±∞ AND g(a) → ±∞.
(ii) You are not applying the quotient rule ! The derivative of the numerator
and denominator are taken separately.
(iii) Sometimes you may find it useful to apply this theorem more than once
to evaluate a limit.
(iv) Sometimes this theorem doesn’t help you evaluate a limit because you get
stuck in a loop between functions.
Let’s have a look at our motivating example again:
Example 2 Evaluate
lim
x→0
3x − sin(x)
.
x
3 − cos(x)
3x − sin(x) LH
=
lim
x→0
x→0
x
1
= 2.
lim
L’Hôpital’s Rule only works on two types of indeterminate forms: 00 and
However, these aren’t the only indeterminate forms we come across. Other
indeterminate forms are:
±∞
∞.
a.
b.
0
0
∞
∞
c.
∞·0
d.
∞−∞
e.
1∞
f.
00
g.
∞0
How do we deal with these other forms? You must find a way to convert them
to one of the two indeterminate forms that L’Hôpital’s Rule works for; forms a.
and b. This can be accomplished by re-writing the limit, putting everything on
a common denominator, taking logarithms of both sides, etc.
2
Example 3 Evaluate
lim xx .
x→0+
Let L = limx→0+ xx . We see that L → 00 . We take logarithms of both sides in
hopes of getting L in a form that we can apply L’Hôpital’s Rule to. Then we
have:
ln(L) = lim+ x ln(x)
x→0
= lim+
x→0
ln(x)
x−1
=LH lim
x→0+
1
x
−x−2
= lim −x
x→0+
= 0.
Since ln(L) = 0, we conclude that L = e0 = 1.
3
Concavity and Curve Sketching
Required Reading: Section 4.4
We have seen how the first derivative gives us information about the graph
of a function (where it is increasing/decreasing and whether a local maximum/minimum occurs at a critical point). We will now see that the second
derivative also gives us important information about the graph of a function
(how the function bends or turns).
Combining the knowledge from the first and second derivative, along with
our previous understanding of symmetry and asymptotes, we will be able to
draw an accurate graph of a function.
When a curve lies above its tangent lines on an interval, we say that the
curve is concave up on that interval. When a curve lies below its tangent lines
on an interval, we say that the curve is concave down on that interval. This can
be re-phrased in terms of the increasing or decreasing of the first derivative:
Definition 1 The graph of a differentiable function y = f (x) is
(a) concave up on an open interval I if f 0 is increasing on I;
(b) concave down on an open interval I if f 0 is decreasing on I.
As we have already learned, when you want to know if something is increasing
or decreasing you study its derivative. Since we are interested in learning when
f 0 is increasing or decreasing, we look at the second derivative:
Theorem 1 (Second Derivative Test for Concavity) Let y = f (x) be twicedifferentiable on an interval I.
1. If f 00 > 0 on I, the graph of f over I is concave up.
2. If f 00 < 0 on I, the graph of f over I is concave down. Example 1 The curve y = x2 is concave up on (−∞, ∞) because its second derivative y 00 = 2 is always positive. Example 2 The curve y = x3 is concave down on (−∞, 0), where y 00 = 6x < 0, and concave up on (0, ∞), where y 00 = 6x > 0.
1
Definition 2 A point (c, f (c)) where the graph of a function has a tangent line
and where the concavity changes (from down to up or from up to down) is a
point of inflection. At a point of inflection (c, f (c)), either f 00 (c) = 0 of f 00 (c)
fails to exist.
We can now combine the different aspects we learned about the behaviour
of a function to formulate a procedure for graphing a function:
Procedure for graphing a function y = f (x)
1. Identify the domain of f and any symmetries the curve may have.
2. Find the derivatives y 0 and y 00 .
3. Find the critical points of f , if any, and identify the function’s behaviour at
each one.
4. Find where the curve is increasing and where it is decreasing.
5. Find the points of inflection, if any occur, and determine the concavity of
the curve.
6. Identify any asymptotes that may exist.
7. Plot key points, such as the intercepts and the points found in steps 3–5, and
sketch the curve together with any asymptotes that may exist.
2
Monotonic Functions and the First Derivative
Test
Required Reading: Section 4.3
When studying the graph of a function, it is useful to know where it increases
(rises as you move from left to right) and where it decreases (falls as you move
from left to right). In this section we learn how to test for intervals of increase
and intervals of decrease using the first derivative. This will also allow us to
identify which local extreme values are present in these intervals.
A function that is increasing only or decreasing only on an interval is said
to be monotonic on that interval. A consequence of the Mean Value Theorem
is that functions with positive derivatives are increasing and functions with
negative derivatives are decreasing:
Corollary 1 Suppose that f is continuous on [a, b] and differentiable on (a, b).
If f 0 (x) > 0 at each point x ∈ (a, b), then f is increasing on [a, b]. If f 0 (x) < 0 at each point x ∈ (a, b), then f is decreasing on [a, b]. It turns out that at the points where the derivative changes sign (positive to negative or negative to positive), i.e. the function changes behaviour from increasing to decreasing or from decreasing to increasing, is where we find our local extreme values: Theorem 1 (First Derivative Test for Local Extrema) Suppose that c is a critical point of a continuous function f , and that f is differentiable at every point in some interval containing c except possibly at c itself. Moving across the interval from left to right: (i) If f 0 changes from negative to positive at c, then f has a local minimum at c. (ii) If f 0 changes from positive to negative at c, then f has a local maximum at c. (iii) If f 0 does not change sign at c, then f has no local extremum at c. Example 1 Find the critical points of f (x) = ex (x2 − 3). 1 Identify the open intervals on which f is increasing and decreasing. Find the function’s local extreme values. The function f is continuous and differentiable over all real numbers x, so the only critical values we will have are when the derivative is equal to zero. f 0 (x) = ex (x2 − 3) + ex (2x) = ex (x2 + 2x − 3) = ex (x + 3)(x − 1). Since ex can never equal zero, our two critical points are x = −3 and x = 1. We test our derivative around these critical points and collect the results in a table: Interval x < −3 −3 < x < 1 x>1
Sign of f 0
+
+
Behaviour of f
Increasing
Decreasing
Increasing
From the table we see that f is increasing on the open interval x ∈ (−∞, −3) ∪
(1, ∞) and decreasing on the open interval x ∈ (−3, 1). We also see that f has
a local maximum of 6e−3 at x = −3 and a local minimum of −2e at x = 1.
2
Homework 5
Week 5 (4.3 – 4.6)
Due August 8, 2020 at 10:00 pm ADT
Read all directions carefully before answering a question. Submit all of your work for every
question. You are welcome to discuss the assignment with other members of the class but
every student must write up their assignment independently.
1. Suppose that f 0 (x) = (x − 7)(x + 1)(x + 5). What are the critical points of f (x)? On what
intervals is f (x) increasing and on what intervals is f (x) decreasing? At what x-values, if any,
does f (x) assume local minimum and local maximum values?
2. Sketch a graph of the rational function
y=
2×2 + x − 1
x2 − 1
3. Evaluate the following limits.
(a)
3x − 1
x→0 2x − 1
lim
(b)
lim x2 ln(x)
x→0+
(c)
lim xx
x→0+
Homework 5
Page 2 of 2
4. Show that among all rectangles with an 8 − m perimeter, the one with largest area is a square.
5. A rectangular plot of farmland will be bounded on one side by a river and on the other three
sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest
area you can enclose, and what are its dimensions?

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