MATH 151 Engineering Math I, Spring 2014
JD Kim
Week4 Section 2.6, 2.7, 3.1, 3.2
Section 2.6 Limits at Infinity; Horizontal Asymptotes
Definition. Let f be a function defined on some interval (a, ∞). Then
lim f (x) = L
x→∞
means that the values of f (x) can be made arbitrarily close to L by taking x sufficiently large.
Definition. Horizontal Asymptote The line y = L is called a horizontal
asymptote of the curve y = f (x) if either
lim f (x) = L or
x→∞
example) limx→∞
x2 − 1
=1
x2 + 1
1
lim f (x) = L.
x→−∞
Ex1) Find the limits:
1-1) limx→∞ x
1-2) limx→−∞ x
1-3) limx→∞ (x − x2 )
1-4) limx→−∞ (x − x3 )
1-5) limx→∞
1
x
1-6) limx→−∞
1
x
1-7) limx→∞
1
x4
1-8) limx→∞
7x3 + 4x
2x3 − x2 + 3
1-9) limt→∞
t4 − t2 + 1
t5 + t3 − t
2
1-10) limx→−∞
1-11) limx→∞
x4 + 2x + 3
x(x2 − 1)
√
1-12) limx→−∞
1-13) limx→−∞
1 + 4x2
4+x
√
1 + 4x2
4+x
√
x2 + 4x
4x + 1
√
1-14) limx→∞ ( x2 + 3x + 1 − x)
1-15) limx→−∞ (x +
√
x2 + 2x)
3
1. Vertical asymptote: undefined point but if it could be cancelled, it
is not vertical asymptote but hole
2. Horizontal asymptote: use infinite limit x → ∞ and x → −∞
Ex2) Find the equation of all vertical and horizontal asymptotes.
2-1) f (x) =
x2
2-2) f (x) = √
x+3
+ 7x + 12
x
x2 + 1
4
Section 2.7 Tangents, Velocities, and other rate
of change
If a curve C has equation y = f (x) and we want to find the tangent to C at
the point p(a, f (a)), then we consider a nearby point Q(x, f (x)), where x 6= a, and
compute the slope of the secant line P Q:
mP Q =
f (x) − f (a)
x−a
Then we let Q approach P along the curve C by letting x approach a. If mP Q
approaches a number m, then we define the TANGENT t to be the line through P
with slope m. (This amounts to saying that the tangent line is the limiting position
of the secant line P Q as Q approaches p.)
Definition. Tangent Line The tangent line to the curve y = f (x) at the point
P (a, f (a)) is the line through P with slope
f (x) − f (a)
x→a
x−a
m = lim
or let x = a + h, then
f (a + h) − f (a)
h→0
h
m = lim
provided that this limit exists.
5
Ex3) Find the slope of the tangent line to the graph of f (x) = x2 + 2x at the
point (1, 3).
Ex4) Find the equation of tangent line to the graph of f (x) =
point is x = a.
Ex5) Find the equation of the tangent line to the graph of f (x) =
6
√
2x + 5 at the
1
at x = 3.
x+2
x+1
. For each vertical
− 2x − 3
asymtote, describe the behavior of f (x) near the asymtote.
Ex6) Find the vertical asymtotes for f (x) =
7
x2
Velocities: Linear motion
If f (t) is the position of an object at tiem t, then
1. The Average Velocity of the object from t = a to t = b is
f (b) − f (a)
∆f
=
∆t
b−a
2. The Instantaneous Velocity of the object at time t = a is
f (a + h) − f (a)
h→0
h
v(a) = lim
Ex6) The position (in meters) of an object moving in a straight path is given by
s(t) = t2 − 8t + 18, where t is measured in seconds.
6-1) Find the average velocity over the time interval [3, 4].
6-2) Find the instantaneous velocity at time t = 3.
8
Other rate of change
Let f (x) be a function
1. The Average rate of change of f (x) from x = a to x = b is
f (b) − f (a)
b−a
2. The Instantaneous Rate of change of f (x) at x = a is
f (a + h) − f (a)
h→0
h
lim
Ex7) If f (x) =
√
x, and
7-1) the average rate of change of f (x) from x = 4 to x = 9.
7-2) the instantaneous rate of change of f (x) at x = 4.
9
Ex8) The population P (in thousands) of a city from 1990 to 1996 is given in the
following table.
Year
Population (×1000)
1990
105
1991
110
1992
117
1993
126
1994
137
1995
150
1996
164
8-1) Find the average rate of growth from 1992 to 1994
8-2) Estimate the instantaneous rate of growth in 1992 by measuring the slope
of a tangent.
10
Ex9) The limit below represents the instantaneous rate of change of some function
f at some number a. State such an f and a:
(2 + h)5 − 32
h→0
h
lim
11
Chapter 3. Derivatives
Section 3.1 Derivatives
Definition. Derivative The Derivative of a function f at a number a, denoted
by f ′ (a), is
f (a + h) − f (a)
f ′ (a) = lim
h→0
h
or
f (x) − f (a)
f ′ (a) = lim
x→a
x−a
if this limit exists.
Ex10) Find the derivative of the following functions at the number a given. (use
definition)
10-1) f (x) = x2 − 8x + 9, a = −2
12
10-2) f (x) =
x
, a=2
x+1
Interpretations of the Derivatives
Geometrically the tangent line to y = f (x) at (a, f (a)) is the line through (a, f (a))
whose slope is equal to f ′ (a), the derivative of f at a.
1. The slope of the tangent line to the graph of f (x) at x = a.
2. The instantaneous rate of change of f (x) at x = a.
3. The instantaneous velocity at x = a.
All use
f (a + h) − f (a)
h→0
h
lim
Ex11) Recall the surface area of a sphere is given by A = 4πr 2 . Find the average
rate of change of the area from r = 1 to r = 2. Find the instantaneous rate of change
of the area at r = 1.
13
Definition. Differentiable
A function f is differentiable at a if f ′ (a) exists.
Theorem.
If f is differentiable at a, then f is continuous at a.
How can a function FAIL to be differentiable
1. If the graph of a function f has a ”corner” or ”kink” in it, then the graph of f
has no tangent at that point and f is not differentiable there.
2. If f is not continuous at a, then f is not differentiable at a.
3. The curve has a vertical tangent line when x = a, f is not differentiable at a.
example) y = |x|.
14
Ex12) The graph of f is given. State, with reasons, the numbers at which f is
not differentiable.
Ex13) Where is f (x) = |x2 − 4| not differentiable.
15
Ex14) Where is f (x) =
|x + 1|
not differentiable.
x+1
Ex15) Given the graph of f (x) below, sketch the graph of the derivative.
15-1)
16
15-2)
Definition. Derivative of function
The derivative of f is defined as
f (x + h) − f (x)
h→0
h
f ′ (x) = lim
17
Ex16) Find the derivative of the following functions as well as the domain of the
derivative.
16-1) f (x) =
3x + 1
x−2
4
16-2) f (x) = √
x
18
Ex17) The limit below represent the derivative of some function f (x) at some
number a. Identify f (x) and a for each limit.
17-1) limh→0
(2 + h)5 − 32
h
17-2) limx→3π
cos x + 1
x − 3π
19