XEQ 201: Calculus II
Chapter 1. Applications of Diﬀerentiation
1.1. Mean value theorems of diﬀerential calculus
1.2. Using diﬀerentials and derivatives
1.3. Extreme Values
Application of diﬀerentiation. Taylor theorem. Mean Value theorem of
diﬀerential calculus. Methods of integration. Applications of integration.
1. Calculus: A complete course by Robert A. Adams and Christopher
2. Fundamental methods of mathematical economics by Alpha C. Chiang.
3. Schaum’s outline series: Introduction to mathematical economics
by Edward T. Dowling
Show that sin x < x for all x > 0.
Solution. If x > 2π, then sin x ≤ 1 < 2π < x. If 0 < x ≤ 2π, then by
MVT, ∃ c ∈ (0, 2π) such that
sin x − sin 0
= [MVT on [0, x]] =
= cos c < 1
which implies that sin x < x in this case too.
Definition 1.1.1 (Increasing and decreasing functions). Suppose that
the function f is deﬁned on an interval I and that x1 and x2 are two points
(a) If f (x2 ) > f (x1 ) whenever x2 > x1 , we say that f is increasing on
(b) If f (x2 ) < f (x1 ) whenever x2 > x1 , we say that f is decreasing on
(c) If f (x2 ) ≥ f (x1 ) whenever x2 > x1 , we say that f is non-decreasing
(d) If f (x2 ) ≤ f (x1 ) whenever x2 > x1 , we say that f is non-increasing
Diagram Fig 2.31
Let J be an open interval and let I be an interval consisting of all points
in J and possibly one or both of the end points of J. Suppose that f is
continuous on I and diﬀerentiable on J.
(x) > 0
(x) < 0
(x) ≥ 0
(x) ≤ 0
x ∈ J,
x ∈ J,
x ∈ J,
x ∈ J,
increasing on I.
decreasing on I.
non-decreasing on I.
non-increasing on I.
On what intervals is the function f (x) = x3 − 12x + 1 increasing? On what
intervals is it decreasing?
Solution. f (x) = 3x2 − 12 = 3 (x − 2) (x + 2). It follows that f (x) >
0 when x < −2 or x > 2 and f (x) < 0 when −2 < x < 2. Therefore f
is increasing on the intervals (−∞, −2) and (2, ∞) and is decreasing on the
interval (−2, 2).
Diagram Fig 2.32
Show that f (x) = x3 is increasing on any interval.
Solution. Let x1 , x2 be any two real numbers satisfying x1 < x2 . Since
f (x) = 3x2 > 0 for all x = 0, we have that f (x1 ) < f (x2 ) if either
x1 < x2 ≤ 0 or 0 ≤ x1 < x2 . If x1 < 0 < x2 , then f (x1 ) < 0 < f (x2 ). Thus
f is increasing on every interval.
If f is continuous on an interval I and f (x) = 0 at every interior point of
I, then f (x) = C, a constant on I.
If f is deﬁned on an open interval (a, b) and achieves a maximum (or a
minimum) at the point c ∈ (a, b), and if f (c) exists, then f (c) = 0.
Values of x where f (x) = 0 are called critical points of the function f .
Theorem 1.1.5 (Rolle’s Theorem).
Suppose that the function g is continuous on the closed ﬁnite interval [a, b]
and if it is diﬀerentiable on the open interval (a, b). If g (a) = g (b), ∃ a
point c ∈ (a, b) such that g (c) = 0.
Theorem 1.1.6 (The Generalized Mean Value Theorem).
If functions f and g are both continuous on [a, b] and diﬀerentiable on (a, b),
and if g (x) = 0 for every x ∈ (a, b), then...