3
Applications of the Derivative
3.1
Increasing and Decreasing Functions
One of our goals is to be able to solve max/min problems, especially economics related
examples.
We start with the following definitions:
Definition 3.1 A function f is called increasing on an interval (a, b) if for any x1 , x2 ∈
(a, b), we have that
x1 < x2 ⇒ f (x1 ) < f (x2 )
A function f is called decreasing on an interval (a, b) if for any x1 , x2 ∈ (a, b), we have
that
x1 < x2 ⇒ f (x1 ) > f (x2 )
Note that some books call these strictly increasing and strictly decreasing respectively.
Examples:
1. f (x) = x2 . f (x) is increasing on (0, ∞) and decreasing (−∞, 0).
2. f (x) = x3 . f (x) is increasing on (−∞, ∞).
As you may have guessed, we can use the derivative to test for increasing/decreasing. Let
f be differentiable on the interval (a, b).
1. If f 0 (x) > 0 for all x in (a, b), then f is increasing on (a, b).
2. If f 0 (x) < 0 for all x in (a, b), then f is decreasing on (a, b).
3. If f 0 (x) = 0 for all x in (a, b), then f must be constant on (a, b).
FACT: If f is a continuous function, then f 0 (x) can only change signs at values of x where
f 0 (x) = 0 or f 0 (x) doesn’t exist.
Examples:
1. Let f (x) =
x3
4
− 3x. Find the intervals on which f (x) is increasing or decreasing.
solution: We will use the test and fact above. The derivative is
f 0 (x) =
3x2
3
3
− 3 = (x2 − 4) = (x − 2)(x + 2)
4
4
4
Because of the above, we split our real line up where f 0 (x) = 0:
−2
f (x)
2
%
&
%
On (−∞, −2), (x − 2) and (x + 2) are both negative, so f 0 (x) = 43 (x − 2)(x + 2) > 0,
so f is increasing.
On (−2, 2), (x − 2) is negative while (x + 2) is positive, so f 0 (x) = 43 (x − 2)(x + 2) < 0,
so f is decreasing.
On (2, ∞), (x − 2) and (x + 2) are both positive, so f 0 (x) = 43 (x − 2)(x + 2) > 0, so f
is increasing.
To summarize, f (x is increasing on (−∞, −2) and (2, ∞) while f (x) is decreasing on
(−2, 2).
2. Let g(x) = x +
32
.
x2
Find the intervals on which g(x) is increasing and decreasing.
solution: We first need to find the derivative.
g 0 (x) = 1 −
64
x3
Note that g 0 (x) = 0 when x3 = 64 ⇒ x = 4. The derivative is also not defined when
x = 0, so these are the places we divide our real line.
0
g(x)
4
%
&
%
To find out what sign the derivative is on (−∞, 0), we can substitute in a test point
in that interval and see if the result is positive or negative. The point x = −1 is a
convenient one to use:
g 0 (−1) = 1 −
64
= 65 > 0
(−1)3
Because of our fact, g 0 (x) will be positive everywhere in (−∞, 0), and thus g is increasing on (−∞, 0).
Similarly,
g 0 (1) = 1 −
64
= −63 < 0
(1)3
So g 0 (x) will be negative everywhere in (0, 4), and thus g is decreasing on (0, 4).
Lastly,
g 0 (5) = 1 −
64
64
>0
=
1
−
(5)3
125
So g 0 (x) will be positive everywhere in (4, ∞), and thus g is increasing on (4, ∞).
3.2
Extrema and the First Derivative Test
We now have enough information to sketch these graphs.
x3
4
1. First let’s graph f (x) =
− 3x. First we find the intercepts.
To get the x-intercept we solve f (x) = 0:
2
x3
x
− 3x = 0 ⇒ x
−3 =0
4
4
√
So the x-intercepts are x = 0, ± 12 ≈ ±3.5. The y-intercept is at (0, 0).
We know that the function changes from increasing to decreasing and vice-versa when
x = ±2. Plugging those values into our function we see that the points (2, −4) and
(−2, 4) are on our graph.
2. Similarly, g(x) = x +
32
.
x2
This function has no y-intercept, because 0 is not in its domain. To find the x-intercept
we solve g(x) = 0:
x+
√
32
3
3
=
0
⇒
x
=
−32
⇒
x
=
−
32 ≈ −3.2
2
x
We have a vertical asymptote at x = 0 because
lim g(x) = +∞ = lim− g(x).
x→0+
x→0
Finally, we calculate the value of g at x = 4 and find that the point (4, 6) is on the
graph.
The point (4, 6) in the above graph is called a relative min.
Definition 3.2 Let f be a function defined at a point c. Then
1. We say that f has a relative minimum (or local minimum) at x = c if there is an
interval (a, b) containing c such that f (c) < f (x) for all x in (a, b).
2. We say that f has a relative maximum (or local maximum) at x = c if there is
an interval (a, b) containing c such that f (c) > f (x) for all x in (a, b).
The word “relative” is significant here. We are not saying that a relative max is largest
value f (x) ever takes. We will get to such points soon.
Together, relative maxes and mins are called relative extrema.
Theorem 3.1 If f (x) has a relative max or min at x = c, then either f 0 (c) = 0 or f 0 (c)
doesn’t exist.
Definition 3.3 If a point c is in the domain of f and either f 0 (c) = 0 or f 0 (c) doesn’t
exist, then we call x = c a critical point or critical value of f .
Examples:
1. f (x) = x2 .
x = 0 is a critical point and a local min. f 0 (0) = 0
2. f (x) = |x|.
x = 0 is a critical point and a local min. f 0 (0) is undefined.
We notice that in the examples above that we have local mins with intervals of negative
slope to the left and positive slope to the right. Under some conditions, this is what happens
in general.
Theorem 3.2 (The First Derivative Test): Let f be continuous on the interval (a, b)
and suppose that c is the only critical point of (a, b). Suppose f is differentiable on (a, b)
except possibly at c. Then:
1. If f 0 (x) is negative to the left of c and positive to the right of c, then f has a local min
at c.
2. If f 0 (x) is positive to the left of c and negative to the right of c, then f has a local max
at c.
3. If f 0 (x) is the same size to the right and left of c then c is neither a max nor a min.
In cases like 3 above we call the point c a saddle point if f 0 (c) = 0.
Examples: Find the critical points and identify their type for the following functions.
1. f (x) = 2x3 + 9x2 − 108x + 30.
solution: We first have to find the critical points
f 0 (x) = 6x2 + 18x − 108 = 6(x2 + 3x − 18) = 6(x + 6)(x − 3)
So the critical points are x = −6, x = 3. To fill in our real line we plug in some test
points from (−∞, −6), (−6, 3) and (3, ∞):
f 0 (−7) = 6(−7 + 6)(−7 − 3) = 60 > 0
f 0 (0) = 6(6)(−3) = −108 < 0
f 0 (4) = 6(4 + 6)(4 − 3) = 60 > 0
Thus we have the following
−6
f 0 (x)
+
f (x)
%
0
3
−
&
0
+
%
So by the first derivative test we can conclude that f (x) has a local max at x = −6
and a local min at x = 3.
2. h(x) = x2 e3x .
solution: We first have to find the critical points:
h0 (x) = 3x2 e3x + 2xe3x = x(3x + 2)e3x
So the critical points are x = 0, x = − 23 . Notice that the e3x doesn’t give us any
critical points because it is defined everywhere and is never 0. To fill in our real line
we plug in some test points from (−∞, − 23 ), (− 23 , 0) and (0, ∞):
h0 (−1) = (−1)(−3 + 2)e−3 = e−3 > 0
1
1
3
1
0
h −
= −
− + 2 e−3/2 = − e−3/2 < 0
2
2
2
4
h0 (1) = (1)(3 + 2)e3 = 5e3 > 0
Thus we have the following
− 23
h0 (x)
+
h(x)
%
0
0
−
&
0
+
%
So by the first derivative test we can conclude that h(x) has a local max at x = − 23
and a local min at x = 0.
3. f (x) = x1/3 ex/6 .
solution: Notice that this function is defined everywhere, i.e. its domain is R. We
once again find the critical points:
1 x/6
1
0
1/3
f (x) = x
e
+ x−2/3 ex/6
6
3
1/3
x
1
x/6
= e
+ 2/3
6
3x
x
2
x/6
= e
+
6x2/3 6x2/3
x+2
x/6
= e
6x2/3
So f 0 (x) = 0 at x = 2 and is undefined at x = 0, and thus we have two critical points.
To fill in our real line we plug in some test points from (−∞, 0), (0, 2) and (2, ∞):
−1
0
−3/6
f (−3) = e
<0
6(−3)2/3
Notice that the denominator is always positive (why?).
1
0
−1/6
f (−1) = e
>0
6(−1)2/3
3
0
1/6
>0
f (1) = e
6(1)2/3
Thus we have the following
−2
f 0 (x)
−
f (x)
&
0
0
+
%
UN
+
%
So there is a local min at x = −2 and at x = 0 there is neither a max nor a min (notice
that there is not a saddle point there because the derivative doesn’t exist there). It is
important to note that x = 0 is still a critical point.
THIS IS A CRUCIAL ISSUE
In the previous example, there was a critical point at x = 0 since f 0 was undefined at x = 0,
but 0 was not in the domain of f .
If x = c is not in the domain, you cannot have a critical point at x = c.
Example: Let f (x) = x1 . Then f 0 (x) = − x12 . f 0 is undefined at x = 0, but there is no critical
point there since f is also undefined at x = 0.
Absolute Extrema
In ecomonics applications, we are generally uninterested in local maxes and mins. Instead
we want to know when our function is maximal or minimal on its entire domain.
Definition 3.4 Let f be defined on an interval I (possibly all of R) containing c. Then f
is said to have a absolute max or global max on I if f (c) ≥ f (x) for every x in I.
There is an analogous statement for absolute/global mins.
The next theorem is both deep and extremely important.
Theorem 3.3 (Extreme Value Theorem): If f is continuous on a closed interval [a, b],
then f has both an absolute max and an absolute min on [a, b].
Note that the word “extreme” in the name of the theorem above is modifying “value” and
not “theorem”. I am not making any claims as to how extreme this theorem is.
The theorem is certainly false for other intervals like R. f (x) = ex has no maxes or mins
on R.
To find extrema on a closed interval [a, b] proceed as follows:
1. Evaluate f at each critical point in [a, b].
2. Evaluate f at the endpoints a and b.
3. The least of these values is the absolute min and the greatest is the absolute max.
Examples:
1. Find the absolute max and min of f (x) = 2x3 − 3x2 − 36x + 2 on the interval [0, 5].
solution: We first find the critical points
f 0 (x) = 6x2 − 6x − 36 = 6(x2 − x − 6) = 6(x − 3)(x + 2)
So the critical points are x = 3 and x = −2. We can throw away x = −2 because it
isn’t in [0, 5]. Now we calculate f at all relevant points.
f (0) = 2
f (3) = 2(27) − 3(9) − 36(3) + 2 = −79
f (5) = 2(125) − 3(25) − 180 + 2 = −3
So the absolute max occurs at x = 0 and the absolute min is at x = 3. Note that the
absolute maxes and mins can occur at endpoints.
2. A company finds that the profit P (x) where x represents thousands of units, is given
by
P (x) = −x3 + 9x2 − 15x − 9
If the company can only make a maximum of 6000 units, what is the absolute maximum
profit?
solution: Since x is in thousands of units, we must find the absolute max of P (x) on
the interval [0, 6].
P 0 (x) = −3x2 + 18x − 15 = −3(x2 − 6x + 5) = −3(x − 5)(x − 1)
So there are two critical points at x = 1, 5. They are both in [0, 6], so we have four
points to calculate:
P (0) = −9
P (1) = −(1)3 + 9(1)2 − 15(1) − 9 = −16
P (5) = −125 + 225 − 75 − 9 = 16
P (6) = −216 + 324 − 90 − 9 = 9
The absolute max occurs at x = 5, so the company should make 5000 units to maximize
profit.
We close out this section with a helpful theorem.
Theorem 3.4 If f is a continuous function on an interval and has only one critical value
in that interval, then a relative max or min is also an absolute max or min.
3.3
Concavity and the Second-Derivative Test
Intuition a curve is concave up on an interval I if it looks like
down on I if it looks like
. We need a more precise definition.
So f 0 is increasing on this interval.
on I. It is concave
And f 0 is decreasing on this interval. These two suggest the following precise definition of
concavity.
Definition 3.5 If f is a function, we say f is concave up on I if f 0 is increasing on I.
We say that f is concave down on I if f 0 is decreasing on I.
Theorem 3.5 Let f be a function. Then
1. If f 00 (x) > 0 for all x in I, then f is concave up on I.
2. If f 00 (x) < 0 for all x in I, then f is concave down on I.
Note that the first graph above has a local min and the second has a local max. This leads
to the following
Theorem 3.6 (The Second-Derivative Test): Suppose that c is a critical point for the
function f . Then
1. If f 00 (c) > 0, then f has a local min at c.
2. If f 00 (c) < 0, then f has a local max at c.
3. If f 00 (c) = 0 or f 00 (c) does not exist, then the test fails.
Note that the second derivative test is easier to use, but sometimes fails. The first derivative
test always works.
Example: Let f (x) =
it is concave down.
6
.
x2 +6
Find all intervals where f is concave up and all intervals where
solution: We have to calculate the second derivative.
f 0 (x) = (−6)(x2 + 3)−2 (2x) =
−12x
(x2 + 3)2
−12(x2 + 3)2 − (−12x)(2(x2 + 3)2x)
(x2 + 3)4
−12(x2 + 3) + 48x2
=
(x2 + 3)3
36x2 − 36
=
(x2 + 3)3
36(x − 1)(x + 1)
=
(x2 + 3)3
f 00 (x) =
The denominator is never 0. f 00 is 0 at x = 1, −1. Thus we can use test points as before to
produce the following diagram:
−1
f 00 (x)
+
0
1
−
0
+
f (x)
It turns out the graph looks like this
Any point at which the concavity of a function changes from up to down or vice-versa is
called a point of inflection.
Theorem 3.7 If c is a point of inflection for f , then either f 00 (c) = 0 or f 00 (x) does not
exist.
Examples:
1. Find the intervals of concavity and points of inflection for f (x) = x4 + x3 − 3x2 + 1.
solution: We calculate the second derivative
f 0 (x) = 4x3 + 3x2 − 6x
f 00 (x) = 12x2 + 6x − 6 = 6(2x − 1)(x + 1)
Thus x = 12 , −1 are our possible inflection points. Thus we can use test points as
before to produce the following diagram:
1
2
−1
f 00 (x)
+
0
−
0
+
f (x)
So f is concave up on (−∞, −1) and ( 21 , ∞) while it is concave down on (−1, 12 ). It
has inflection points at both x = 12 and x = −1.
Note that just because f 00 (c) = 0 does not mean c is an inflection point (exercise: show
that f (x) = x4 has a point where the second derivative is zero but x = c is not an
inflection point).
2. Use the second derivative test to find and classify the critical points for
f (x) = (x2 − 8x + 16)ex
.
solution: We calculate the first and second derivatives:
f 0 (x) = (2x − 8)ex + (x2 − 8x + 16)ex = (x2 − 6x + 8)ex = (x − 2)(x − 4)ex
f 00 (x) = (x2 − 6x + 8)ex + (2x − 6)ex = (x2 − 4x + 2)ex
When x = 4, f 00 (x) = 2ex > 0. So there is a local min at x = 4.
When x = 2, f 00 (x) = −2ex < 0. So there is a local max at x = 2.
3.4
Word Problems
We can use calculus to solve real-world problems! Yay!
Examples:
1. We wish to enclose with a fence a rectangular region of 12,250 square feet next to a
wall. Only 3 sides need to be fenced. The fencing for one side perpendicular to the
wall costs $ 30/ft. The other two sides cost $ 20/ft. What are the dimensions of the
rectangle that minimize cost? What is the minimum cost?
solution:
WALL
$30/ft
x
x
$20/ft
y
$20/ft
So the total cost is C = 30x + 20y + 20x = 50x + 20y.
This is a two variable relation. We use the area stipulation to reduce it to one variable.
A = 12, 250 = xy ⇒ y =
12, 250
x
So what we would like to minimize is
245, 000
x
We use the derivative to find the minimum of this function
245, 000
C 0 (x) = 50 −
x2
C(x) = 50x +
Note that 0 is not a critical point because C is undefined there. We solve C(x) = 0
50 = 245, 000x− 2
x2 =
245, 000
= 4900
50
x = 70
In this problem x must be positive so we don’t consider the negative square root. So
we have a critical point at x = 70. Now
C 00 (x) =
490, 000
x3
490, 000
>0
703
So C(x) has a local minimum at x = 70. By the theorem at the end of last class
(Theorem 3.4 in these notes), since it is the only critical point it must be an absolute
minimum. Thus
12, 250
= 175
y=
70
So the minimum cost occurs when the dimensions are 70ft by 175ft. The minimum
cost is
245, 000
C(70) = 50(70) +
= 3500 + 3500 = $7000
70
C 00 (70) =
2. (Old exam question)
A 150 room hotel will rent all of its rooms if it charges $40 per room. For each $2
increase in price, 3 fewer rooms will be rented per night. Find a price that maximizes
revenue.
solution: Let x = number of rooms rented, and let p(x) = price per room (this is a
demand function).
So, total revenue is R(x) = xp(x).
We notice that p(x) must be linear, since each increase in price leads to a constant
change in p(x)
x p(x)
150
40
147
42
44
144
141
46
If p(x) is linear p(x) = mx + b. We can find m easily:
m=
∆y
40 − 42
2
=
=−
∆x
150 − 147
3
So y = − 32 x + b. We plug in a point:
2
40 = − (150) + b ⇒ b = 140
3
Thus p(x) = − 23 x + 140, and so R(x) = xp(x) = − 23 x2 + 140x. This is what we would
like to maximize.
4
R0 (x) = − x + 140
3
3
So our only critical point is x = − 4 (−140) = 105 rooms. So the maximum revenue is
obtained when we rent out 105 rooms which corresponds to a price of
2
p(105) = − (105) + 140 = $70.
3
Why is this an absolute max? We could appeal to theorem 3.4 again, but we could also
notice that R(x) is a quadratic polynomial with a negative x2 coefficient and therefore
its graph is a downwards parabola.
The end of this last problem brings up an important point - when doing these types of
problems you have to verify that your solution is an absolute max or min. There are three
ways of doing this:
1. If the function is a simple one like a parabola you may be able to verify it by inspection.
For example, parabolas have the form f (x) = ax2 + bx + c and have only one critical
point. If a > 0 it is an absolute min and if a < 0 it is an absolute max.
2. We can use this theorem: If f is continuous on a closed interval [a, b], then f has both
an absolute max and an absolute min on [a, b]. To find them,
(a) Evaluate f at each critical point in [a, b].
(b) Evaluate f at the endpoints a and b.
(c) The least of these values is the absolute min and the greatest is the absolute max.
3. We can use this theorem: If f is a continuous function on an interval and has only one
critical value in that interval, then a relative max or min is also an absolute max or
min.
Example:
A manufacturer wants to design an open box with a square base and a surface area of 108
in2 . What dimensions will maximize volume? What is the maximum volume?
solution:
h
x
x
We are given that the surface area is fixed at 108 in2 . This allows us to solve for h in terms
of x:
SA = x2 + 4hx
108 = x2 + 4hx
108 − x2
h =
4x
2
We want to maximize V = x h. As a function of x it is
V (x) = hx2
108 − x2
=
x2
4x
108x − x3
=
4
We find the critical points
1
V 0 (x) = (108 − 3x2 ) = 0 ⇒ x2 = 36 ⇒ x = ±6
4
We only consider the critical point x = 6 because here x is a physical dimension and
negative numbers here don’t make sense (our domain is all positive numbers, (0, ∞)).
The second derivative is V 00 (x) = − 6x
, so
4
V 00 (6) = −9 < 0
and by the second-derivative test the critical point x = 6 is a local max. Since x = 6 is the
only critical point on the interval (0, ∞), by our theorem above x = 6 must be an absolute
max on that interval.
The height that maximizes volume is
108 − 36
108 − x2
=
=3
h=
4x
24
and so the maximum volume is
V = x2 h = (6)(6)(3) = 108 in3
3.5
More Economics Applications
Price Elasticity of Demand
One way economists measure the responsiveness of consumers to a change in the price of
a product is with what is called price elasticity of demand. For example, changing the
price on vegetables usually strongly affects the demand (we call products like this elastic)
while changing the price of milk or water doesn’t affect that demand that much (we call
products like this inelastic).
Definition 3.6 If p(x) is a differentiable demand function, then the price elasticity of
demand is given by
p(x)/x
η=
dp/dx
where η the lowercase Greek letter eta. For a given price, the demand is said to be elastic
if |η| > 1 and the demand is said to be inelastic if |η| < 1. The demand is unit elasticity
if |η| = 1.
For example, the text gives the elasticity for some common commodities as:
Tomatoes
Automobiles
Housing
Mail
η
η
η
η
= 4.60
= 1.35
= 1.00
= 0.05
Example:
Let the demand function for a product be modelled by p(x) = 21 −
elasticity of demand when x = 36 and x = 400.
solution: Here
dp
dx
= − 4√3 x . So for x = 36:
p(36) = 21 −
3p
(36)
2
= 12
3
p0 (36) = − √
4 36
1
= −
8
Thus
η=
12/36
8
=−
−1/8
3
Here |η| > 1 so the demand is elastic.
For x = 400:
p(400) = 21 −
3p
(400)
2
= −9
3
p0 (400) = − √
4 400
3
= −
80
Thus
η=
Here |η| < 1 so the demand is inelastic.
−9/400
3
=
−3/80
5
3√
x.
2
Find the price
Average Cost
Due to the nature of cost functions, the minimum cost usually corresponds to making 0
units. This isn’t very useful information, as it is obvious that if we make no units we
will spend the least money. More informative would be to find the production level that
minimizes the cost per unit, or average cost:
C̄(x) =
C(x)
x
Example:
Suppose that our cost function is C(x) = 800 + 0.04x + 0.0002x2 . Find the production level
that minimizes the average cost.
solution: In this case the averave cost is
C̄(x) =
C(x)
800
=
+ 0.04 + 0.0002x
x
x
We find the critical points:
C̄ 0 (x) = −
800
800
2
+
0.0002
=
0
⇒
x
=
= 4,000,000
x2
0.0002
Since we are again only considering x in (0, ∞), we have one critical point, x = 2000. Now,
C̄ 00 (x) =
1600
x3
1600
>0
20003
so x = 2000 is a local minimum, and that together with the fact that it is the only critical
point in the domain (0, ∞) gives us that it is an absolute minimum. The minumum average
cost is
800
C̄(2000) =
+ 0.04 + 0.0002(2000) = $0.84
2000
C̄ 00 (2000) =
3.6
Asymptotes
The function f (x) =
3
x−2
has a graph that looks like this
Here we say that x = 2 is a vertical asymptote for the graph of f . Notice that
lim f (x) = ∞
x→2+
lim f (x) = −∞
x→2−
This leads to the following definition.
Definition 3.7 If f (x) approaches ∞ or −∞ as x approaches some fixed c from the right
or left, then the line x = c is said to be a vertical asymptote of the graph of f .
If f is a rational function, ie
f (x) =
g(x)
h(x)
then c is a vertical asymptote for f if h(x) = 0 and g(x) 6= 0.
Examples:
1. Suppose that f is defined as
f (x) =
x+4
x+4
=
2
x − 4x
x(x − 4)
Then the denominator is 0 when x = 0 or x = 4. Neither of these make the denominator 0, so both x = 0 and x = 4 are vertical asymptotes. To find out the behaviour
of f near the asymptotes we compute the one-sided limits:
x+4
+
lim f (x) = lim−
x→0−
x→0 x(x − 4)
(−)(−)
= +∞
x+4
+
lim f (x) = lim+
x→0+
x→0 x(x − 4)
(+)(−)
= −∞
+
x+4
lim f (x) = lim−
x→4−
x→4 x(x − 4)
(+)(−)
= −∞
x+4
+
lim f (x) = lim+
x→4+
x→4 x(x − 4)
(+)(+)
= +∞
We summarize this with the following graph
2. Suppose that f is defined as
f (x) =
x2 + 2x − 8
(x − 2)(x + 4)
=
2
x −4
(x − 2)(x + 2)
In this case we have a vertical asymptote at x = −2 but not at x = 2. Note that f is
still not defined at the point x = 2 even though the (x − 2) term cancels out. When
x 6= 2, we can cancel
(x − 2)(x + 4)
(x − 2)(x + 2)
x+4
=
x 6= 2
x+2
x+2+2
=
x 6= 2
x+2
2
= 1+
x 6= 2
x+2
f (x) =
2
So we see that everywhere except x = 2, f looks like 1 + x+2
. At x = 2, f has a hole.
Notice that the last function appears to approach a value as x goes to ∞. This leads to
the following definition.
Definition 3.8 If f is a function and L is a real number, the statements
lim f (x) = L or
x→∞
lim f (x) = L
x→−∞
denote limits at ∞. In either of these cases, the line y = L is called a horizontal asymptote for f .
Examples:
1. For any r > 0,
1
=0
x→∞ xr
2. In the last example with the graph above,
lim
f (x) =
x2 + 2x − 8
2
=
1
+
for x 6= 2
x2 − 4
x+2
2
If we take the limit as x approaches ∞, the x+2
term gets smaller and smaller, and so
2
=1
lim 1 +
x→∞
x+2
So the function has a horizontal asymptote at y = 1.
A common situation is finding horizontal asymptotes for rational functions p(x)
. There are
q(x)
a few simple rules that will let you easily see what asymptotes such functions have.
3. Find the horizontal asymptotes of f (x) =
2x+1
.
4x2 +5
solution: To evaluate this limit, we multiply the top and bottom by the reciprocal of
the largest power of x involved (in this case, that will be x12 ):
1 2x + 1
2x + 1
x2
lim
= lim
1
2
x→∞ 4x2 + 5
x→∞
4x + 5
x2
+ x12
x→∞ 4 + 52
x
0+0
=
4+0
= 0
=
lim
2
x
So f (x) has a horizontal asymptote at y = 0.
4. Find the horizontal asymptotes of g(x) =
2x2 +1
4x2 +5
solution: We can use the same trick as before:
2
1 2x2 + 1
2x + 1
x2
lim
= lim
1
2
x→∞ 4x2 + 5
x→∞
4x + 5
x2
2+
x→∞ 4 +
2+0
=
4+0
1
=
2
=
lim
1
x2
5
x2
So g(x) has a horizontal asymptote at y = 12 .
The last couple examples can be summarized as follows: suppose f (x) =
q are both polynomials. Then
p(x)
,
q(x)
where p and
• If the degree of p is less than the degree of q, then f has a horizontal asymptote at
y = 0.
• If the degree of p is equal to the degree of q, then f has a horizontal asymptote at
y = L where L is the ratio of the leading coefficient of p to the leading coefficient of q.
• If the degree of p is greater than the degree of q, then the function has no horizontal
asymptotes.
Example: (Real life example of asymptotic behaviour)
A small business invests $5000 in a new product. In addition to the initial investment, the
product costs $0.50 per unit to produce. Find the average cost per unit
a) if 1000 units are produced,
b) if 10000 units are produced,
c) as the number of units produced goes to infinity.
solution: We can just read the formula off and evaluate at 1000 and 10000 to get a) and b)
x
2
5000 1
C̄(x) =
+
x
2
1
C̄(1000) = 5 + = $5.50
2
1 1
C̄(10000) =
+ = $1
2 2
C(x) = 5000 +
To get c) we evaluate C̄(x) as x → ∞:
lim C̄(x) = 0 +
x→∞
1
= $0.50
2
This makes sense; as you make more and more units the cost due to the initial investment
becomes less and less significant.
3.7
Curve Sketching
This section is a summary of the information we have gained in this chapter and how to
apply it to sketching the graph of a given function. When asked to graph a function, your
graph should take into account the following information:
• x-intercepts and y-intercept,
• domain,
• continuity (ie places where it is not continuous),
• differentiability (ie places where it isn’t differentiable),
• local maxes and mins,
• intervals of increase and decrease,
• intervals of concavity,
• points of inflection,
• asymptotes.
Examples:
1. Sketch the graph of y = f (x) = x3 + 3x2 − 9x − 11.
solution: We try to find all the information in the list above
• To find the y-intercept we find f (0) = −11. So the y-intercept is (0, −11).
The x-intercepts in this example are a bit tricky to find (and you are not expected
to do this from scratch). More often I will give you one of the roots and you will
have to find the other one. Here, one of the roots is x = −1. Thus (x + 1) has to
be a factor of f (x), and you can find by long division (or any other method) that
f (x) = (x + 1)(x2 + 2x − 11)
So by using the quadratic formula we find that the other roots are
p
√
−2 ± 4 − 4(−11)
= −1 ± 12 ≈ 2.5 and − 4.5
x=
2
The domain of f is all real numbers.
• To find the critical points and intervals of increase/decrease we need the derivative
f 0 (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1)
So there are two critical points at x = −3 and x = 1. Using the first derivative
test we get
−3
f 0 (x)
+
f (x)
%
0
1
−
&
0
+
%
So there is a local max at x = −3 and a local min at x = 1. f is increasing
on (−∞, −3) and (1, ∞) while it is decreasing on (−3, 1). The local max is at
(−3, 16) and the local min is at (1, −16).
• To find inflection points and intervals of concavity we need the second derivative
f 00 (x) = 6x + 6
This is positive when x > −1 and negative when x < −1. So there is an inflection
point at x = −1; plugging this into the function gives us that the inflection point
is at (−1, 0). In addition, f is concave up on (−1, ∞) and concave down on
(−∞, −1).
We now have all the information we need to graph this function
2. Sketch the graph of g(x) =
x+2
.
x−1
solution: Here the asymptotes are crucial, so we’ll tackle them first.
• There is a vertical asymptote at x = 1 since the denominator is zero there but the
top is not.
lim − g(x) = ∞ lim+ g(x) = −∞
x→dir0o
x→1
• There is a horizontal asymptote at y = 1 since g is a rational function with the
same degree on the top and bottom and the ratio of leading coeffiecients is 11 = 1.
• The x-intercept is at x = −2 and the y-intercept is at
all real numbers excluding x = 1.
2
−1
= −1. The domain is
• We calculate the derivative
g 0 (x) =
1(x − 2) − 1(x + 1)
−3
=
2
(x − 1)
(x − 1)2
This is never 0. It is undefined at x = 1, but since x = 1 is not in the domain, it
is not a critical point. Also notice that g 0 (x) is always negative when x 6= 1, so g
is decreasing on (−∞, 1) and (1, ∞).
• We calculate the second derivative:
g 00 (x) =
6
(x − 1)3
This is undefined at x = 1, but so is g so g has no inflection points. g 00 (x) > 0
when x > 1 and g 00 (x) < 0 when x < 1. So g is concave down on (−∞, 1) and
concave up on (1, ∞).
So our graph looks like this:
3. Sketch the graph of
f (x) =
x2 + x − 6
(x − 2)(x + 3)
=
2
x + x − 30
(x − 5)(x + 6)
solution: The asymptotes play an important part here as well, so again we’ll get to
them first.
• f has vertical asymptotes at x = 5 and at x = −6. Near the asymptotes we have
lim f (x) = −∞
x→5−
lim f (x) = ∞
x→−6−
lim f (x) = ∞
x→5+
lim f (x) = −∞
x→−6+
• Since this is a rational function with equal degrees on top and bottom, we can see
that f has a horizontal asymptote equal to the ratio of the leading coefficients, ie
y = 1.
• The domain of f is all real numbers except 5 and −6. It has x-intercepts at x = 2
and x = −3 and y-intercept at 15 .
• One can use the quotient rule to find the derivative to be
f 0 (x) =
−24(2x + 1)
(x − 5)2 (x + 6)2
Thus there is a critical point at x = − 21 . Note once again that 5 and −6 are not
critical points because they are not in the domain of f .
− 21
−6
f 0 (x)
+
f (x)
%
UN
+
%
0
5
−
&
UN
−
&
So from the above we can see that there is a local max at x = − 21 . Calculating
25
25
f (− 12 ) = 121
, we see the local max is at the point (− 12 , 121
).
• Exercise: show that this function has no inflection points.
This last point tells us that we don’t need any concavity information to graph this
function. So the graph looks like this: