Calculus I
Mrs. Farber
What is Calculus?
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Two Basic Problems of Calculus
1. Find the slope of the curve y = f (x) at the
point (x, f (x))
(x, f(x))
(x, f(x))


 (x, f(x))
2. Find the area of the region bounded above by
the
curve y = f(x), below by the x-axis and by the
Area
vertical lines x = a and x = b
y = f(x)
a
b
x
From BC (before calculus)
We can calculate the slope of a line given two points
change in y y2  y1 y
slope 


change in x x2  x1 x
Calculate the slope of the line
between the given point P (.5, .5)
and another point on the curve,
say Q(.1, .99). The line is called
a secant line.
.99  .5 .49
slope 

 1.225
.1  .5 .4
Slope of Secant line PQ
P(0.5, 0.5)
Point Q
x f(x)
.1 .99
.2 .98
.3 .92
.4 .76
Let x values get closer and closer to .5. Determine f(x) values.
Slope of Secant line PQ
As Q gets closer to P, the
Slope of the secant line PQ
Gets closer and closer to
the slope
Of the line tangent to the
Curve at P.
Figure 1.4: The tangent line at point P has the same steepness
(slope)
that the curve
at P.
Slope
of has
a curve
at a point
The slope of the
curve at a point P is
defined to be the
slope of the line
that is tangent to
the curve at point P.
In the figure the
point is P(0.5, 0.5)
Slope formula
change in y y2  y1 y
slope 


change in x x2  x1 x
In calculus we learn how to calculate the slope at a given point P.
The strategy is to take use secant lines with a second point Q.
and find the slope of the secant line.
Continue by choosing second points Q that are closer and closer
to the given point P and see if the difference quotient gets closer
to some fixed value.
Find the slope of y = x2 at the point (1,1) Find the equation of
the tangent line.
Slope
Q approaches P from the right
Q approaches P from the left
A
Find slope of tangent line on f(x) =x2 at the point (1,1)
Approaching x = 1 from the right
x
f(x)
2
4
Slope of secant
between (1,1)
and (x, f(x))
3
1.5
2.25
2.5
1.1
1.21
2.1
1.01
1.021
2.01
1.001
1.002001
2.001
Slope appears to be getting close to 2.
Find slope of tangent line on f(x) =x2 at the point (1,1)
Approaching x = 1 from the left
x
f(x)
0
0
Slope of secant
between (1,1)
and (x, f(x))
1
.5
.25
1.5
.9
.81
1.9
.99
.9801
1.99
.999
.998001
1.999
Slope appears to be getting close to 2.
Write the equation of tangent line
As the x value of the second point gets
closer and closer to 1, the slope gets
closer and closer to 2. We say the limit of
the slopes of the secant is 2. This is the
slope of the tangent line.
To write the equation of the tangent line
use the point-slope formula
y  y1  m( x  x1 )
y  1  2( x  1)
y  2x 1
bc)
change in y y2  y1 y
Average rate of change 


change in x x2  x1 x
If f(t) represents the position of an object as a function of time,
then the rate of change is the velocity of the object.
Find the average velocity if f (t) = 2 + cost on [0, ]
1. Calculate the function
value (position) at each
endpoint of the interval
f() = 2 + cos () = 2 – 1 = 1
f(0) = 2 + cos (0) = 2 + 1 = 3
y2  y1 y 1  3 2



t2  t1
t   0 
2
 .6366
The average velocity on on [0, ] is
2. Use the slope formula

Instantaneous rate of change
To calculate the instantaneous rate of change of
we could not use the slope formula since we do
not have two points.
To approximate instantaneous calculate the
average rates of change in shorter and shorter
intervals to approximate the instantaneous rate
of change.
2.2
To understand the
instantaneous rate of
change (slope) problem and
the area problem, you will
need to learn about
limits
Limits
x 8
f ( x) 
x2
3
What happens to the value of f (x) when the value of x
gets closer and closer and closer (but not necessarily
equal) to 2?
x 8
x2
3
We write this as:
lim x2
The answer can be found graphically,
numerically and analytically.
Graphical Analysis
20
18
16
14
12
10
8
6
4
2
f (x)
5
4
x 8
x2
3
lim x2
3
2
21
4
0
1
2
3
4
What happens to f(x)
as x gets closer to 2?
5
x
Numerical Analysis lim
Use one sided limits
x 2
x3  8
x2
x 8
x2
3
lim x2
Start to the left of 2 and choose x values getting closer and
closer (but not equal) to 2
x
1.5
1.9
1.99
f (x)
9.25
11.41
11.941
1.999
11.994001
1.9999
11.99940001
Could x get closer to 2? Does f(x) appear to get
closer to a fixed number?
Numerical Analysis lim
x 8
x2
3
lim x2
x 2
x3  8
x2
Start to the right of 2 and choose x values
getting closer and closer (but not equal ) to 2
x
2.5
2.1
2.01
f (x)
15.3
12.61
12.0601
2.001
12.006001
2.0001
12.00060001
If the limit exists, f(x) must approach the same
value from both directions. Does the limit exist?
Guess what it is.
Figure 1.8: The functions in Example 7.
Limits that do not exist
In order for a limit to exist, the function must approach the same value
From the left and from the right.
Infinite Limits
x2
lim
x 3 x  3
What happens to the function value as x gets
closer and closer to 3 from the right?
x 3.5 3.1 3.01 3.001 3.0001 3.00001 3.000001
y 3
11
101
1001
10001
The function increases without
bound so we say
x2
lim

x 3 x  3
There is a vertical asymptote at x = 3.
100001 1000001
51
41
31
21
11
1
9
19
29
39
49
The line x=a is a Vertical Asymptote if
at least one is true.
lim f ( x)  
lim f ( x)  
lim f ( x)  
lim f ( x)  
xa
xa
xa
xa
Identify any vertical asymptotes:
x2
f ( x) 
x5
x2
f ( x)  2
x  5x  6
x2
f ( x)  2
x  5x  6
x2
f ( x)  2
x 6
Graph of f(x)
x  7  6.999  1.80
t  2.2  2.205  7
True or false
(a) x = 2 is in the domain of f
(b) lim exists
x 2
(c) lim x 2 f ( x)  lim x 2 f ( x)
2.3 Functions That Agree at All But One Point
If f(x) = g(x) for all x in an open interval except x = c then:
Example
lim xc f ( x )  lim xc g ( x )
x 2  7 x  10
 x 5
x2
if x  2
then
lim
x 2
x 2  7 x  10
 lim
x2
x 2
( x  5)
Evaluate by direct substitution 2-5 = -3
As x gets closer and closer and closer to 2, the function
value gets closer and closer to -3.
Analytic
=
=
lim
x 2
lim
x 8
x2
3
lim x2
( x  2)( x 2  2 x  4)
x2
x 2
(x
2
 2 x  4)
Using direct substitution,
2 2  2(2)  4  12
As x gets closer and closer to 2 (but not equal to 2)
f(x) gets closer and closer to 12
Compute some limits
x 4
x2
2
lim x 2
x 8
x2
lim x 3
x3  8
x2
3
lim x 0
lim x 2
x3  8
x2
Basic Limits
If b and c are real numbers and is n a positive integer
1.
lim xc b  b
lim x7  2
Ex:
2.
lim xc x  c
Ex:
3.
lim xc x  c
Ex:
=5
lim x5 x
n
= -2
n
lim x3 x
2
=9
Guess
Guess
Guess
an an
an
answer
answer
answer
click
andand
and
click
click
to to
to
check.
check.
check.
Properties of Limits
Multiplication by a constant b
lim xa bf ( x)  b lim xa f ( x)
Limit of a sum or difference
lim xa  f ( x)  g ( x)  lim xa f ( x)  lim xa g ( x )
Limit of a product
lim xa  f ( x) * g ( x)  lim xa f ( x) * lim xa g ( x)
Limit of a power
lim xa  f ( x)  lim xa f ( x)
n
Limit of a quotient
f ( x ) lim xa f ( x )
lim xa

, lim xa g ( x )  0
when denominator
g ( x ) lim xa g ( x )
is not 0.
n
Using Properties of Limits
Properties allow evaluation of limits by direct substitution for
many functions.
Ex.:
3x 2 ( x  2) lim x3 3x 2 ( x  2)
lim x3

x6
lim x3 x  6
lim x3 3x 2 * lim x3 ( x  2)
lim x3 ( x  6)
3(lim x3 x )2 * lim x3 ( x  2)
lim x3 ( x  6)
3(3) 2 * (3  2) 3(9)(1)

9
3
( 3  6)
As x gets closer and closer to 3, the function value gets
closer and closer to 9.
Analytic Techniques
Direct substitution
First substitute the value of x being
approached into the function f(x). If this is
a real number then the limit is that number.
If the function is piecewise defined, you must
perform the substitution from both sides of x.
The limit exists if both sides yield the same
value. If different values are produced, we say
the limit does not exist.
Analytic Techniques
Rewrite algebraically if direct substitution
produces an indeterminate form such
as 0/0
Factor and reduce
Rationalize a numerator or denominator
Simplify a complex fraction
When you rewrite you are often producing another function that
agrees with the original in all but one point. When this happens
the limits at that point are equal.
Find the indicated limit
0
direct substitution fails
x  x6
lim
x 3
x3
2
0
( x  3)( x  2)
lim
x 3
x3
lim ( x  2)
x 3
=-5
Rewrite and cancel
now use direct sub.
Find the indicated limit
lim
x 0
lim
x 0
x 1 1
x
0
direct substitution fails
0
x 1 1
x 1 1
x
*
 lim
x
x  1  1 x 0 x[ x  1  1]
Rewrite and cancel
lim
x 0
1
1

x 1 1 2
now use direct sub.
Find the indicated limit
5

 x 2  1, x  2
f ( x)  

5 x  3, x  2
lim f ( x)
calculate one sided limits
7
x 2
lim f ( x)  5
x 2
lim f ( x)  7
x 2
Since the one-sided limits are not equal,
we say the limit does not exist. There will
be a jump in the graph at x =2
Figure 1.24: The graph of f () = (sin )/.
Determine the limit on y =
sin θ/θ as θ approaches 0.
Although the function is not defined at
θ =0, the limit as θ
0 is 1.
Figure 1.37: The graph of y = e1/x for x < 0 shows
limx0– e1/x = 0. A
(Example
11)
one-sided
limit
limx0 0
Limits that are infinite
(y increases without bound)
lim x 4
1

x4
lim x 2
1
 
2 x
lim x  3
1
 
x3
An infinite limit will exist as x approaches a finite value when
direct substitution produces
not zero
0
If an infinite limit occurs at x = c we have a vertical
asymptote with the equation x = c.
FigureContinuity
1.50: The function
in (a)at
is continuous
at xnot
= 0; the
2.5
in (a)
x = 0 but
in
functionsgraphs.
in (b) through ( f ) are not.
other
Conditions for continuity
A function y = f(x) is continuous at x = c if and
only if:
• The
function is defined at x = c
• The limit as x approaches c exists
• The value of the function and the value of
the limit are equal.
f (c)  lim f ( x)
x c
Find the reasons for
discontinuity in b, c, d, e
and f.
Figure 1.53: Composites of continuous functions are
continuous.
Composite Functions
If two functions are continuous at x = c then their composition
will be continuous.
Example:
f ( x)  x  4
2
is continuous for all reals.
Exploring Continuity
cx
if x  1

if x  1
4
 x 3  mx if x  1

2
Are there values of c and m that make the function continuous
At x = 1? Find c and m or tell why they do not exist.
Exploring Continuity
lim f ( x)  c(1)2  c
cx
if x  1
x 1

f
(1)

4
4
if
x

1

 x 3  mx if x  1 lim f ( x)  (1)3  m(1)  1  m

2
x 1
c4
1  m  4
m5
2.6 Slope of secant line
and slope of tangent line
y f ( x)  f (a)
msec 

x
xa
mtan  lim x a
f ( x)  f (a )
xa
s(t) = 8(t3 – 6 t2 +12t)
Position of a car at t hours.
1. Draw a graph.
t
0
1
2
3
s
0
56
64
72
2. Does the car ever stop?
3. What is the average velocity for the following intervals
a. [0, 2],
b. [.5, 1.5]
c. [.9,1.1]
4. Estimate the instantaneous velocity at t = 1
s(t) =
8(t3
–6
t2
+12t)
2. Appears to stop at t =2. (Velocity= 0)
3. What is the average velocity for
[0, 2],
[.5, 1.5]
[.9,1.1]
t
s(t)
0
0
2
62
.5
37
1.5
63
.9
53.352
1.1
58.168
80
70
60
50
40
30
20
10
0 0.5
1
1.5
2
a) 31 mph
b) 26 mph
c) 24.08 mph
2.5
3
Find an equation of the tangent line
to y = 2x3 – 4 at the point P(2, 12)
f ( x)  f (a )
mtan  lim x a
xa
(2 x3  4)  12
2 x3  16
lim x 2
 lim x 2
x2
x2
2( x  2)( x 2  2 x  4)
lim x 2

x2
lim x2 2( x2  2 x  4)  24
So, m = 24. Use the point slope form to write the equation
y  12  24( x  2)
y  24 x  36
Figure 1.62: The tangent slope is
lim f (x0 + h) – f (x0)
h0
h
Slope of the tangent line at x= a
Q(a + h, f (a + h))
f(a+h) – f(a)
P(a, f(a))
a
a+h
Other form for Slope of
secant line of tangent line
Let h = x - a
Then x = a + h
y f (a  h)  f (a)
msec 

x
h
mtan  limh0
f ( a  h)  f ( a )
h
Find an equation of the tangent line at
2
(3, ½) to y 
x 1
f ( a  h)  f ( a )
mtan  limh0
h
2
2
2(a  1)  2(a  h  1)

limh0

lim h 0 a  h  1 a  1
h(a  1)(a  h  1)
h
2a  2  2a  2h  2
2h
limh0
 limh0
h(a  1)(a  h  1)
h(a  1)(a  h  1)
2
2
lim h 0

At a = 3, m = - 1/8
(a  1)(a  h  1) (a  1) 2
Using the point-slope formula:
1
1
y    ( x  3)
2
8
1
7
y x
8
8