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Functions, Limits, and the Derivative
2
• Functions and Their Graphs
• The Algebra of Functions
• Functions and Mathematical Models
• Limits
• One-Sided Limits and Continuity
• The Derivative
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Function
A rule that assigns to each element in a set A (the
domain), one and only one element in a set B (the
range)
Range
Domain
-1
1
3
-4
1
-6
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Function Notation
y  3 x 2  2 is a function, with values of x
as the domain and values of y as the range.
We write f ( x) in place of y.
This is read “f of x.”
2
f
(
x
)

3
x
2
So
NOTE: It is not f times x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Function Notation
Example
Find f (2), f (0), and f (b  1), where f ( x)  3 x 2  2.
Solution
f (2)  3(2)  2  3(4)  2  14
2
Plug in –2
f (0)  3(0) 2  2  2
f (b  1)  3(b  1) 2  2
 3(b 2  2b  1)  2  3b 2  6b  5
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Domain of a Function
The domain of a function is the set of values for x
for which f (x) is a real number.
2x
Ex. Find the domain of f ( x) 
3x  1
Since division by zero is undefined we must
1
have 3 x  1  0  x  .
3
The domain can be expressed as the intervals
1

1 

,
and


 ,
3

3 
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Domain of a Function
Ex. Find the domain of f ( x)  7 x  5
Since the square root of a negative number is
undefined we must have
5
7 x  5  0  7x  5  x   .
7
5
The domain is all values of x that satisfy x   .
7
The domain can be expressed as the interval
 5 
  ,
 7 
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Graph of a Function
The graph of a function is the set of all points (x, y)
such that x is in the domain of f and y = f (x).
Given the graph of y = f (x),
find f (1).
y
f (1) = 2
(1, 2)
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Graph of a Function
Vertical Line Test: The graph of a function can be
crossed at most once by any vertical line.
Function
Not a Function
It is crossed
more than
once.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Sketch the graph of the function:
 2x  5
f ( x)  
 x  7
x  2
x  2
.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Algebra of Functions
 f  g  x   f ( x)  g ( x)
 f  g  x   f ( x)  g ( x)
 fg  x   f ( x) g ( x)
 f 
f ( x)
  x 
g ( x)
g


Domain:
Domain of f intersected
with the domain of g.
Domain:
Domain of f intersected
with the domain of g with
the exclusion of all values
of x, such that g(x) = 0.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Given f ( x)  3x  1 and g ( x)  x  2 x  3
2
2
f

g
x

3
x

1

x
 2x  3

 
 x  x4
2
 f  g  x   3x  1  ( x
 2 x  3)
  x 2  5x  2
 fg  x    3x  1  x
2
2
 2x  3

 3x3  7 x2  7 x  3
 f 
3x  1
2
, x  2x  3  0
   x  2
x  2x  3
g


Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Composition of Functions
 f g  x   f ( g ( x))
2
Ex. Given f ( x) 
, g ( x)  x3  3 find
3x  5
 f g  ( x) and  g f  (1).

3
f
g
(
x
)

f
g
(
x
)

f
x
3


 
2


3
3
3
x
 14
3 x 3 5

g
2


f  (1)  g  f (1)   g  1  4
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Types of Functions
Polynomial Functions
P( x)  an x  an1x
n
n1
 ...  a0
 an  0
n is a nonnegative integer, each ai is a constant.
1 4
3
f
(
x
)

x

x
 12 x  5
Ex.
2
Rational Functions
F ( x) 
f  x
g ( x)
polynomials
Ex. F ( x)  3x  4 2
2x  7 x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Types of Functions
Power Functions
f ( x)  x
r
( r is any real number)
Ex.
g ( x)  x
Ex.
1
1/ 3
f ( x)  3  x
x
2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Application Example 1 (Functions )
A shirt producer has a fixed monthly cost of $5000.
If each shirt costs $3 and sells for $12 find:
a.
The cost function
Cost: C(x) = 3x + 5000 where x is the number
of shirts produced.
b.
The revenue function
Revenue: R(x) = 12x where x is the number of
shirts sold.
c.
The profit from 900 shirts
Profit: P(x) = Revenue – Cost
= 12x – (3x + 5000) = 9x – 5000
P(900) = 9(900) – 5000 = 3100, or $3100.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Application Example 2 (Functions)
A division of Chapman Corporation manufactures a
pager. The weekly fixed cost for the division is
$20,000, and the variable cost for producing x
pagers/week is
V ( x)  0.000001x3  0.01x 2  50 x dollars.
The company realizes a revenue of
R( x)  0.02 x 2  150 x  0  x  7500  dollars
from the sale of x pagers/week.
......
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
1. Find the total cost function.
The total cost function is the variable cost plus the
fixed cost:
3
2
C ( x)  0.000001x  0.01x  50 x  20,000
2. Find the total profit function.
The profit is the revenue minus the total cost
P( x)  0.02 x  150 x 
2

0.000001x3  0.01x 2  50 x  20,000

P( x)  0.000001x3  0.01x 2  100 x  20, 000
......
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
3. What is the profit for the company if 2000 units
are produced and sold each week?
Since the profit function is
P( x)  0.000001x3  0.01x 2  100 x  20,000
we have
P(2000) 
0.000001(2000)3  0.01(2000)2  100(2000)  20,000
 132,000, or $132,000
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Introduction to Calculus
There are two main areas of focus:
1. Finding the tangent line to a curve at a given
y
point.
y  f ( x)
 x1, y1 
x
tangent line
2. Finding the area of a planar region bounded by
a given curve.
y
Area
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Velocity
Average  Distance covered
Elapsed time
Over any time
interval
If I travel 200 miles in 5 hours, my average
velocity is 40 miles/hour.
Distance covered As elapsed time
Instantaneous 
Elapsed time
approaches zero
When I see the police officer, my instantaneous
velocity is 60 miles/hour.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Velocity
Ex. Given the position function s  t   t 2  10t
where t is in seconds and s(t) is measured in
feet, find:
a.
The average velocity for t = 1 to t = 3.
Velocityave
b.
Notice how
elapsed time
approaches zero
s (3)  s(1) 39  11


 14 ft/sec
3 1
2
The instantaneous velocity at t = 1.

t
1.1
1.01
1.001
Average velocity
12.1
12.01
12.001
s (t )  s (1)
t 1
Answer: 12 ft/sec
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Limit of a Function
The limit of f (x), as x approaches a, equals L
written: lim f ( x)  L
xa
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to a.
y
y  f ( x)
L
a
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Computing Limits
3x if x  2
Ex. lim f ( x) where f ( x)  
x 2
1 if x  2
y
lim f ( x) = lim  3x
x2
6
 3 lim x
x 2
 3(2)  6
Note: f (-2) = 1
is not involved
x2
x
-2
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Properties of Limits
Suppose lim f ( x)  L and lim g ( x)  M
xa
xa
Then,
1. lim  f ( x)   Lr
r
xa
r , a real number
2. lim cf ( x)  c lim f ( x)  cL c, a real number
xa
xa
3. lim  f ( x)  g ( x)   L  M
x a
4. lim  f ( x) g ( x)   LM
xa
f ( x)
f ( x) lim
L
xa
5. lim


x a g ( x)
lim g ( x) M
Provided that M  0
xa
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Computing Limits
Ex. lim  x  1  lim x  lim1
2
2
x3
x 3
x 3
   lim1
 lim x
2
x 3
x 3
 32  1  10
lim  2 x  1
2 lim x  lim1
2x 1
x 1
x 1
x 1
Ex. lim


x 1 3 x  5
lim  3 x  5 
3lim x  lim 5
x 1
x 1
x 1
2 1 1


35 8
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Indeterminate Form:
x5
Ex. xlim
5 x 2  25
0
0
0
Notice form
0
x5
 lim
x5  x  5  x  5 
Factor and cancel
common factors
1
1
 lim

x 5  x  5 
10
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Limits at Infinity
1
1
For all n > 0, lim n  lim n  0
x x
x x
1
provided that n is defined.
x
5 1
 2
2
3x  5 x  1
Divide
x
x
 lim
Ex. xlim
2
2
2
x


by x
2  4x

4
x2
5
 1 
lim 3  lim    lim  2 
x 
x   x  x   x 
3 0 0
3



 2 
04
4
lim  2   lim 4
x   x  x 
3
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
One-Sided Limit of a Function
The right-hand limit of f (x), as x approaches a,
equals L
written: lim f ( x)  L
x a
if we can make the value f (x) arbitrarily close
to L by taking x to be sufficiently close to the
right of a.
y  f ( x)
L
a
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
One-Sided Limit of a Function
The left-hand limit of f (x), as x approaches a,
equals M
written: lim f ( x)  M
x a
if we can make the value f (x) arbitrarily close
to M by taking x to be sufficiently close to the
y
left of a.
y  f ( x)
M
a
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
One-Sided Limit of a Function
 x 2 if x  3
Ex. Given f ( x)  
2x if x  3
Find lim f ( x)
x3
lim f ( x)  lim 2 x  6
x 3
x 3
Find lim f ( x)
x3
lim f ( x)  lim x2  9
x3
x3
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Continuity of a Function
A function f is continuous at the point x = a if
the following are true:
i ) f (a ) is defined
y
ii) lim f ( x) exists
xa
iii) lim f ( x)  f (a)
xa
f(a)
a
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Properties of Continuous Functions
The constant function f (x) is continuous
everywhere.
Ex. f (x) = 10 is continuous everywhere.
The identity function f (x) = x is continuous
everywhere.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Properties of Continuous Functions
If f and g are continuous at x = a, then
f
f  g , fg , and
g
at x  a.
 g (a)  0  are continuous
A polynomial function y = P(x) is continuous at
everywhere.
A rational function
R( x) 
p ( x)
q ( x)
is continuous
at all x values in its domain.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Intermediate Value Theorem
If f is a continuous function on a closed interval [a, b]
and L is any number between f (a) and f (b), then there
is at least one number c in [a, b] such that f(c) = L.
y
y  f ( x)
f (b)
f (c) = L
f (a)
x
a c
b
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Intermediate Value Theorem
Ex.
Given f ( x)  3 x 2  2 x  5. Show that f ( x)  0
has at least one solution on 1, 2  .
f (1)  4  0 and f (2)  3  0
f (x) is continuous for all values of x and since
f (1) < 0 and f (2) > 0, by the Intermediate Value
Theorem, there exists a c on (1, 2) such that
f (c) = 0.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Existence of Zeros of a Continuous
Function
If f is a continuous function on a closed interval [a, b],
and f(a) and f(b) have opposite signs, then there is at least
one solution of the equation f(x) = 0 in the interval (a, b).
y
f(b)
a
b
x
f(a)
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
(Existence of zeros of a continuous function)
Let f ( x)  x  3x  5.
2
1. Show that f(x) is a continuous function everywhere.
The function is a polynomial function and is
therefore continuous everywhere.
2. Show that f(x) = 0 has at least one solution on the
interval (0, 2)
Since f (0)  5 and f (2)  5 have opposite signs,
there must be at least one number x  c with
0  c  2 such that f (c)  0.
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Rates of Change
Average rate of change of f over the interval
[x, x+h]
f ( x  h)  f ( x )

Slope of Secant Line
h
Instantaneous rate of change of f at x
f ( x  h)  f ( x )
 lim
h 0
h
Slope of the
Tangent Line
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
The Derivative
The derivative of a function f with respect to x is
the function f , given by
f ( x  h)  f ( x )
f ( x)  lim
h 0
h
It is read “f prime of x.”
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
The Derivative
Four-step process for finding f  :
1. Compute f ( x  h)
2. Find
f ( x  h)  f ( x )
f ( x  h)  f ( x )
h
f ( x  h)  f ( x )
4. Compute f ( x)  lim
h 0
h
3. Find
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
The Derivative
Given f ( x)  2 x 2  1, find f ( x).
1. f ( x  h)  2  x  h   1  2 x  4 xh  2h  1
2
2
2
2
2
2
f
(
x

h
)

f
(
x
)

2
x

4
xh

2
h

1

(2
x
 1)
2.
 4 xh  2h2
f ( x  h)  f ( x) 4 xh  2h 2
3.

h
h
2
f
(
x

h
)

f
(
x
)
4
xh

2
h
4. lim
 lim
h 0
h 0
h
h
 lim  4 x  2h   4 x  f ( x)  4 x
h 0
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
Find the slope of the tangent line to the graph of
f ( x)  2 x  3 at any point (x, f(x)).
Step 1.
f ( x  h)  2( x  h)  3  2 x  2h  3
Step 2.
f ( x  h)  f ( x)  (2 x  2h  3)  (2 x  3)  2h
Step 3.
f ( x  h)  f ( x) 2h

 2
h
h
Step 4.
f ( x  h)  f ( x )
f ( x)  lim
 lim  2   2
h 0
h 0
h
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Differentiability and Continuity
If a function is differentiable at x = a, then it is
continuous at x = a.
y
Not
Continuous
x
Not
Differentiable
Still
Continuous
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example
The function f ( x)  x is not differentiable at
x = 0 but it is continuous everywhere.
y
f ( x)  x
O
x
Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc.