Download Sections 1.10-1.11

(Section 1.10: Difference Quotients) 1.10.1
SECTION 1.10: DIFFERENCE QUOTIENTS
LEARNING OBJECTIVES
• Define average rate of change (and average velocity) algebraically and graphically.
• Be able to identify, construct, and evaluate (and simplify) various forms of
difference quotients.
PART A: DISCUSSION
• This section will revisit Section 0.14 on slopes of lines, Section 1.1 on function
evaluations, and Section 1.2 on graphs.
• A difference quotient is used to find the slope of a secant line to a graph or to find
an average rate of change (perhaps an average velocity). Difference quotients will
form the basis for derivatives, tangent lines, and instantaneous rates of change in
Section 1.11.
PART B: SECANT LINES and AVERAGE RATE OF CHANGE
The secant line to the graph of a function f on the interval a, b , where a < b ,
( ( )) and ( b, f ( b)) .
is the line that passes through the points a, f a
The average rate of change of f on a, b is equal to the slope of this secant line,
rise f b f a
=
which is given by:
.
b a
run
()
()
We call this a difference quotient, because it has the form:
• (See Footnote 1 on assumptions about f .)
difference of outputs
.
difference of inputs
(Section 1.10: Difference Quotients) 1.10.2
PART C: AVERAGE VELOCITY
The following development of average velocity will help explain the association
between slope and average rate of change.
Example 1 (Average Velocity)
A car is driven due north 100 miles during a two-hour trip.
What is the average velocity of the car?
• Let t = the time (in hours) elapsed since the beginning of the trip.
()
• Let y = s t , where s is the position function for the car (in miles).
s gives the signed distance of the car from the starting position.
•• The position (s) values would be negative if the car were south of
the starting position.
()
• Let s 0 = 0 , meaning that y = 0 corresponds to the starting position.
()
Therefore, s 2 = 100 (miles).
The average velocity on the time-interval a, b is the
average rate of change of position with respect to time. That is,
change in position s
=
change in time
t
where (uppercase delta) denotes “change in”
=
( ) ( ) , a difference quotient
s b s a
b a
Here, the average velocity on 0, 2 is:
( ) ( ) = 100 0
s 2 s 0
20
2
miles mi
= 50
or
or
mph
hour hr
TIP 1: The unit of velocity is the unit of slope given by:
unit of s
.
unit of t
(Section 1.10: Difference Quotients) 1.10.3
The average velocity is 50 mph on 0, 2 in the three scenarios below.
It is the slope of the orange secant line.
We will define instantaneous velocity (or simply velocity) in Section 1.11.
• Here, the velocity is constant (50 mph).
• Here, the velocity is increasing; the car is accelerating.
• Here, the car overshoots the destination and then backtracks.
WARNING 1: The car’s velocity is negative in value when it is
backtracking; this happens when the graph falls.
• In calculus, the Mean Value Theorem for Derivatives will imply that the car must be
going exactly 50 mph at some time value t in 0, 2 . The theorem applies in all three
( )
( )
scenarios above, because s is continuous on 0, 2 and is differentiable on 0, 2 ,
meaning that its graph makes no sharp turns and does not exhibit “infinite steepness” on
0, 2 . Differentiability will be discussed in Section 1.11. §
( )
(Section 1.10: Difference Quotients) 1.10.4
PART D: FORMS OF DIFFERENCE QUOTIENTS
The purposes of these forms will be discussed in Section 1.11 on derivatives.
Forms of Difference Quotients
Form 1: Fixed interval
()
()
f b f a
b a
a, b are constants
Form 2: Variable endpoint (x)
()
()
f x f a
xa
a is constant;
x is variable
Form 3: Variable run (h)
(
)
()
f a+h f a
h
a is constant;
h is variable
Form 4: Variable endpoint (x) and Variable run (h)
(
)
()
f x+h f x
x, h are variable
h
• The denominators can be negative. (See Footnote 2 in Section 1.11.)
(Section 1.10: Difference Quotients) 1.10.5
PART E: EVALUATING DIFFERENCE QUOTIENTS
Example 2 (Form 1 of a Difference Quotient; Profit)
A company sells widgets. Assume that all widgets produced are sold.
Let P be the profit function for the company; P x is the profit (in dollars)
()
if x widgets are produced and sold. Our model: P ( x ) = x
2
+ 200x 5000 .
Find the average rate of change of profit between 60 and 90 widgets.
)
WARNING 2: We will treat the domain of P as 0, , even though one
could argue that the domain should only consist of integers. Be aware of this
issue with applications such as these.
§ Solution
( )
( )
( )
( )
90 2 + 200 90 5000 60 2 + 200 60 5000 P 90 P 60
=
30
90 60
( )
( )
WARNING 3: Grouping symbols are essential when
expanding P 60 here, since we are subtracting an
( )
expression with more than one term.
8100 + 18,000 5000 3600 + 12,000 5000 =
30
8100 + 18,000 5000 + 3600 12,000 + 5000
=
30
( )
TIP 2: Instead of completely evaluating P 90 and
( )
P 60 , it may help to take advantage of cancellations
such as the one above.
1500
30
dollars
= 50
widget
=
(Section 1.10: Difference Quotients) 1.10.6
• The practical interpretation is that, if the company increases its production
level from 60 widgets to 90 widgets, then its profit will increase by
dollars
, on average.
50
widget
• The total increase in profit is $1500 over the 30 additional widgets.
()
• The graph of y = P x is below.
•• The slope of the orange secant line is also 50
dollars
.
widget
•• The fact that the slope is positive reflects the fact that the profit
function is increasing on 60, 90 . We will revisit this idea in
Section 1.11 on derivatives. §
(Section 1.10: Difference Quotients) 1.10.7
Example 3 (Form 2 of a Difference Quotient; Revisiting Example 2 on Profit)
()
Again, P x = x 2 + 200x 5000 . Evaluate
()
( ) . (Form 2)
P x P 60
x 60
§ Solution
()
( )=
P x P 60
x 60
( )
( )
2
x 2 + 200x 5000 60 + 200 60 5000 x 60
x 2 + 200x 5000 3600 + 12,000 5000 =
x 60
x 2 + 200x 5000 + 3600 12,000 + 5000
=
x 60
2
x + 200x 8400
=
x 60
x 2 200x + 8400
=
x 60
(
)
• The Factor Theorem in Chapter 2 will imply that,
if P is polynomial, then x 60 is a factor of the
(
)
()
( )
numerator, which is equivalent to P x P 60 .
This is because 60 is a zero of the numerator:
P 60 P 60 = 0 .
( )
=
(
( )
) ( x 140)
, ( x 60 )
x
60
(
)
x 60
1
(
) ( x 60)
( x 60)
= x 140 ,
= 140 x,
(Section 1.10: Difference Quotients) 1.10.8
dollars
, though we tend to omit
widget
it when a variable is present in our final expression.
• The unit for the difference quotient is still
dollars • If we substitute x = 90 , we obtain 50 , our answer to Example 2.
widget §
(Section 1.10: Difference Quotients) 1.10.9
Example 4 (Form 3 of a Difference Quotient; Revisiting Example 2 on Profit)
()
Again, P x = x 2 + 200x 5000 . Evaluate
(
)
( ) . (Form 3)
P 60 + h P 60
h
§ Solution
(
)
( )
P 60 + h P 60
h
(
)
(
)
( )
( )
60 + h 2 + 200 60 + h 5000 60 2 + 200 60 5000 =
h
3600 + 120h + h2 + 12,000 + 200h 5000 3600 + 12,000 5000 =
h
2
3600 120h h + 12,000 + 200h 5000 + 3600 12,000 + 5000
=
h
2
80h h
=
h
h 80 h
=
, h0
h
(
)
(
)
(
)
1
= 80 h,
( h 0)
• If we substitute h = 30 , which corresponds to x = 90 , we obtain
dollars 50 , our answer to Example 2. §
widget (Section 1.10: Difference Quotients) 1.10.10
Example 5 (Form 4 of a Difference Quotient; Revisiting Example 2 on Profit)
()
Again, P x = x 2 + 200x 5000 . Evaluate
(
)
( ) . (Form 4)
P x+h P x
h
§ Solution
(
)
()
P x+h P x
(
h
)
(
)
x + h 2 + 200 x + h 5000 x 2 + 200x 5000 =
h
x 2 + 2xh + h2 + 200x + 200h 5000 x 2 + 200x 5000 =
h
2
2
x 2xh h + 200x + 200h 5000 x 2 + 200x 5000 =
h
(
=
=
=
)
x 2 2xh h2 + 200x + 200h 5000 + x 2 200x + 5000
h
2xh h2 + 200h
h
h 2x h + 200
(
),
h
( h 0)
1
= 2x h + 200,
( h 0)
• If we substitute x = 60 and h = 30 (which then corresponds to x = 90 ),
dollars we obtain 50 , our answer to Example 2. §
widget (Section 1.10: Difference Quotients) 1.10.11
FOOTNOTES
1. Assumptions made about a function. When defining the average rate of change of a
function f on an interval a, b , where a < b , sources typically do not state the assumptions
()
()
f b f a
()
seems only to require the existence of f a and
b a
f b , but we typically assume more than just that.
made about f . The formula
()
• Although the slope of the secant line on a, b can still be defined, we need more for
the existence of derivatives (i.e., the differentiability of f ) and the existence of
non-vertical tangent lines, as we will see in Section 1.11.
• We ordinarily assume that f is continuous on a, b . Then, there are no holes, jumps,
or vertical asymptotes on a, b when f is graphed. (See Section 1.5.)
• We may also assume that f is differentiable on a, b . Then, the graph of f makes no
sharp turns and does not exhibit “infinite steepness” (corresponding to vertical tangent
lines). However, this assumption may lead to circular reasoning, because the ideas of
secant lines and average rate of change are used to develop the ideas of derivatives,
tangent lines, and instantaneous rate of change, as we will see in Section 1.11.
Differentiability is defined in terms of the existence of derivatives.
• We may also need to assume that f , the derivative of f , is continuous on a, b .
Then, the average rate of change of f on a, b is equal to the average value of f on
a, b . In calculus, we will assume that a function (say, g) is continuous on a, b and
g ( x ) dx ; the numerator is a
b
then define the average value of g on a, b to be
a
b a
definite integral, which will be defined as a limit of sums in calculus (see Section 9.8).
Then, the average value of f on a, b is given by:
()
()
f b f a
b
a
()
f x dx
b a
, which is equal to
by the Fundamental Theorem of Calculus. The theorem assumes that the
b a
integrand [function], f , is continuous on a, b .
(Section 1.11: Limits and Derivatives in Calculus) 1.11.1
SECTION 1.11: LIMITS AND DERIVATIVES IN CALCULUS
LEARNING OBJECTIVES
• Be able to develop limit definitions of derivatives and use them to find derivatives.
• Understand the graphical interpretation of a derivative as the slope of a tangent line.
• Understand the practical interpretation of a derivative as an instantaneous rate of
change, possibly velocity.
• Be able to find equations of tangent lines to graphs.
• Relate derivatives to local linearization of, and marginal change in, a function.
• Use derivatives to determine where a function is increasing, decreasing, or constant.
PART A: DISCUSSION
• In Section 1.5, we discussed limits. In Section 1.10, we saw how difference
quotients represented slopes of secant lines and average rates of change.
• We will now define derivatives as limits of difference quotients. Derivatives
represent slopes of tangent lines (which model or approximate graphs locally) and
instantaneous rates of change.
• Difference quotients on “small” intervals might be used to approximate
derivatives. Conversely, derivatives might be used to approximate marginal change
in a function, an idea used in economics.
• In calculus, we will use derivatives to help us graph functions by finding where
they are increasing, decreasing, or constant. (See Section 1.2.)
• Limits, differentiation (the process of taking derivatives), and integration
(which reverses differentiation; see Section 9.8) are the three key topics you will
find in the first half of a calculus book, and they will be key themes throughout.
(Section 1.11: Limits and Derivatives in Calculus) 1.11.2
PART B: TANGENT LINES and DERIVATIVES
In Section 1.10, we looked at four forms of difference quotients.
Form 1 gave us the average rate of change of a function f on a fixed interval
a, b , where a < b . This is equal to the slope of the secant line on a, b .
Form 1: Fixed interval
()
()
f b f a
a, b are constants
b a
The other forms dealt with variable intervals. Each form, together with the idea of
limits from Section 1.5, leads to a definition of derivative.
Form 2: Variable endpoint (x)
()
()
f x f a
a is constant;
x is variable
xa
(
)
• If we let x approach a denoted by: x a , the corresponding secant lines
approach the red tangent line in the following figures.
• This “limiting process” makes the tangent line a creature of calculus, not
just precalculus. (See Footnote 1 on etymology.)
(Section 1.11: Limits and Derivatives in Calculus) 1.11.3
Below, we let x approach a
from the right x a + .
(
Below, we let x approach a
from the left x a .
)
(
)
(See Footnote 2.)
• We define the slope of the tangent line to be the (two-sided) limit of the
difference quotient as x approaches a, if that limit exists.
()
• We denote this slope by f a , read as “ f prime of (or at) a.”
()
f a , the derivative of f at a, is the slope of the tangent line to the graph
( ( ))
of f at the point a, f a , if that slope exists (as a real number).
\
()
• If f a exists, we then say that f is differentiable at a.
Limit Definition of the Derivative at a Point (Based on Form 2)
()
f a = lim
xa
()
()
f x f a
xa
• The term “point” here technically refers to the real number a, which
corresponds to a point on the real number line.
(Section 1.11: Limits and Derivatives in Calculus) 1.11.4
Form 3: Variable run (h)
(
)
()
f a+h f a
a is constant;
h is variable
h
(
)
• If we let the run h approach 0 denoted by: h 0 , the corresponding
secant lines approach the red tangent line below.
Below, we let h approach 0
from the right h 0+ .
(
Below, we let h approach 0
from the left h 0 .
)
(
)
(See Footnote 2.)
Limit Definition of the Derivative at a Point (Based on Form 3)
()
f a = lim
h0
(
)
()
f a+h f a
h
(Section 1.11: Limits and Derivatives in Calculus) 1.11.5
Form 4: Variable endpoint (x) and Variable run (h)
(
)
()
f x+h f x
x, h are variable
h
• Again, if we let h 0 , the corresponding secant lines approach the red
tangent line below.
• The figures are similar to the ones for Form 3, except that x replaces a.
Below, we let h approach 0
from the right h 0+ .
(
Below, we let h approach 0
from the left h 0 .
)
(
)
(See Footnote 2.)
()
• f x provides a rule for f , the derivative function (or simply derivative)
of f . TIP 1: Think of f as a slope function.
Limit Definition of the Derivative Function (Based on Form 4)
()
f x = lim
h0
(
)
()
f x+h f x
h
(Section 1.11: Limits and Derivatives in Calculus) 1.11.6
PART C: INSTANTANEOUS RATE OF CHANGE and
INSTANTANEOUS VELOCITY
()
The instantaneous rate of change of f at a is equal to f a , if it exists.
The following development of instantaneous velocity will help explain the
association between the slope of a tangent line and instantaneous rate of change.
Example 1 (Velocity)
A car is driven due north for two hours, beginning at noon.
How can we find the instantaneous velocity of the car at 1pm?
(If this is positive, this can be thought of as the speedometer reading
at 1pm.)
Re Example 1: Definitions
• Let t = the time (in hours) elapsed since noon.
• Let y = s t , where s is the position function for the car (in miles).
()
In Section 1.10, we saw that:
The average velocity on the time-interval a, b is the
average rate of change of position with respect to time. That is,
change in position s
=
change in time
t
=
() ()
s b s a
b a
We will now consider average velocities on variable time intervals of the
form a, a + h , if h > 0 , or the form a + h, a , if h < 0 ,
where h is a variable run. (We can let h = t .)
(Section 1.11: Limits and Derivatives in Calculus) 1.11.7
Using Form 3 of a difference quotient,
The average velocity on the time-interval a, a + h , if h > 0 , or
a + h, a , if h < 0 , is given by:
(
) ()
s a+h s a
h
• This equals the slope of the secant line to the graph of s on the interval.
• (See Footnote 2 on the h < 0 case.)
Let’s assume there exists a non-vertical tangent line to the graph of s at the
point a, s a .
( ( ))
• Then, as h 0 , the slopes of the secant lines will approach the
slope of this tangent line, which is s a .
()
• Likewise, as h 0 , the average velocities will approach the
instantaneous velocity at a.
Below, we let h 0+ .
Below, we let h 0 .
The instantaneous velocity (or simply velocity) at a is given by:
()
()
s a , or v a = lim
h0
(
) ()
s a+h s a
h
(Section 1.11: Limits and Derivatives in Calculus) 1.11.8
Using Form 4 of a difference quotient,
()
The velocity function v is defined by: v t = lim
(
) ()
s t+h s t
h0
h
Re Example 1: Numerical approaches; Approximating derivatives
()
Let’s say the position function s is defined by: s t = t 3 on 0, 2 .
We want to find v 1 , the instantaneous velocity of the car at t = 1 or a = 1.
()
We will first consider average velocities on intervals of the form 1, 1 + h .
Here, we let h 0+ .
Interval
Value of h
(in hours)
1, 2 1
1, 1.1
0.1
1, 1.01
0.01
1, 1.001
0.001
Average velocity,
(
) ()
s 1+ h s 1
h
( ) ( ) = 7 mph
s 2 s 1
1
s 1.1 s 1
( ) ( ) = 3.31 mph
0.1
s 1.01 s 1
(
) ( ) = 3.0301 mph
0.01
s 1.001 s 1
(
) ( ) = 3.003001 mph
0.001
3 mph
0+
()
• These average velocities approach 3 mph, which is v 1 .
()
The reader can verify that v 1 = 3 mph in the Exercises.
Part D will help.
• WARNING 1: Tables can sometimes be misleading. The table here
does not represent a rigorous evaluation of v 1 . Answers are not
()
always integer-valued.
• If h is “close” to 0, we may be able to use the corresponding average
velocity as an approximation for v 1 .
()
(Section 1.11: Limits and Derivatives in Calculus) 1.11.9
We could also consider this approach:
Average velocity,
(
) ()
s 1+ h s 1
Interval
Value of h
h
(rounded off to six significant digits)
1, 2 1 hour
7.00000 mph
1 minute
3.05028 mph
1 second
3.00083 mph
0+
3 mph
1
1, 1 60 1 1, 1 3600 Here, we let h 0 .
Interval
Value of h
(in hours)
0, 1
1
0.9, 1
0.1
0.99, 1
0.01
0.999, 1
0.001
0
Average velocity,
(
) ()
s 1+ h s 1
h
( ) ( ) = 1 mph
s 0 s 1
1
( ) ( ) = 2.71 mph
s 0.9 s 1
(
0.1
) ( ) = 2.9701 mph
s 0.99 s 1
(
0.01
) ( ) = 2.997001 mph
s 0.999 s 1
0.001
3 mph
• Because of the way we normally look at slopes, we may prefer
s 1 s 0
s 0 s 1
to rewrite the difference quotient
as
,
1
1
and so forth. (See Footnote 2.) §
( ) ()
() ( )
(Section 1.11: Limits and Derivatives in Calculus) 1.11.10
PART D: FINDING DERIVATIVES
Example 2 (Profit; Revisiting Examples 2-5 in Section 1.10)
A company sells widgets. Assume that all widgets produced are sold.
Let P be the profit function for the company; P x is the profit (in dollars)
()
if x widgets are produced and sold. Our model: P ( x ) = x
2
+ 200x 5000 .
Find the instantaneous rate of change of profit at 60 widgets.
• We will ignore integer restrictions on x.
§ Solution
( )
We want to find P 60 . We will present three solutions based on three
different forms of difference quotients we developed in Section 1.10.
Form 2
( )
P 60 = lim
()
( )
P x P 60
x 60
= lim 140 x ,
x 60
x 60
(
) ( x 60) ( by Ex.3 in Section 1.10)
• As x approaches 60, we see that 140 x approaches
140 60 , or 80. We can substitute x = 60 directly; the
( )
(
)
restriction x 60 is irrelevant here.
( )
= 140 60
= 80
dollars
widget
• This is the slope of the red tangent line below.
(Section 1.11: Limits and Derivatives in Calculus) 1.11.11
Form 3
( )
P 60 = lim
(
)
( )
P 60 + h P 60
h0
h
(
) ( h 0) ( by Ex.4 in Section 1.10)
= 80 ( 0 )
= lim 80 h ,
h0
= 80
dollars
widget
Form 4
()
P ( x + h) P ( x )
First, find the formula for P x , the rule for the derivative of P.
()
P x = lim
h0
h
= lim 2x h + 200 ,
h0
(
()
) ( h 0) ( by Ex.5 in Section 1.10)
= 2x 0 + 200
= 2x + 200
( )
Now, evaluate P 60 .
( )
( )
P 60 = 2 60 + 200
= 80
dollars
widget
Form 3
Form 4
§
(Section 1.11: Limits and Derivatives in Calculus) 1.11.12
PART E: EQUATIONS OF TANGENT LINES
Example 3 (Profit; Revisiting Example 2)
()
Again, P x = x 2 + 200x 5000 . Find Slope-Intercept and Point-Slope
()
Forms of the equation of the tangent line to the graph of y = P x at the
(
( ))
point 60, P 60 . (Review Section 0.14 on these forms.)
§ Solution
( )
P ( 60 ) = ( 60 ) + 200 ( 60 ) 5000
= 3400 ( dollars )
• Find P ( 60 ) , the slope of the desired tangent line.
• Find P 60 , the y-coordinate of the desired point.
2
In Part D, we showed (three times) that:
dollars P 60 = 80 widget ( )
• Find a Point-Slope Form of the equation of the tangent line.
( )
y 3400 = 80 ( x 60 )
y y1 = m x x1
• Find the Slope-Intercept Form of the equation of the tangent line.
y = 80x 1400
• Note: Some sources would use P instead of y as the dependent variable.
§
(Section 1.11: Limits and Derivatives in Calculus) 1.11.13
PART F: LINEARIZATION and MARGINAL CHANGE
( ( )) represents
• The tangent line to the graph of a function f at the point a, f a
the best linear approximation to the function close to a. The tangent line model
linearizes the function locally around a.
• The principle of “local linearity” states that, if the tangent line exists, the graph
of f resembles the line if we “zoom in” on the point a, f a .
( ( ))
Example 4 (Marginal Profit; Revisiting Examples 2 and 3)
If P is a profit function, the derivative P is referred to as a marginal profit
(MP) function. It approximates the change in profit if we increase
production by 1 unit (from 60 to 61 widgets, for instance). That is,
P 60 P 61 P 60 .
( )
( )
( )
• This is based on the idea that the tangent line to the graph of P at
60, P 60 represents the best linear approximation to P “close to”
(
( ))
x = 60 (or a = 60 ).
( )
( )
• P 61 P 60 is the actual rise along the graph of P as x changes from 60
to 61. We will approximate this by the rise along the tangent line as x
changes from 60 to 61. This rise is given by P 60 , the slope of the tangent
( )
line, because, if run = 1, slope =
rise
= rise . (We saw this in Section 0.14.)
run
• (The blue graph of P below has been distorted for convenience. The true
graph is very close to the red tangent line.)
§
(Section 1.11: Limits and Derivatives in Calculus) 1.11.14
PART G: GRAPHING FUNCTIONS USING DERIVATIVES
Example 5 (Profit; Revisiting Example 2)
()
Again, P x = x 2 + 200x 5000 . Graph P.
§ Solution
We will discuss how to graph quadratic functions such as P in Section 2.2.
For now, we can use the derivative function P to help us graph P.
()
• In Part D (Example 2, Form 4), we found that P x = 2x + 200 .
Think of this as a “slope function” that gives us slopes of tangent lines.
• Find any point(s) on the graph where the tangent line is horizontal;
its slope is 0.
Solve:
()
P x = 0
2x + 200 = 0
x = 100
(
( ))
There is a horizontal tangent line at the point 100, P 100 , or
(100, 5000) .
( )
( )
y-coordinate of the point. What is P (100 ) ?
WARNING 2: Evaluate P 100 , not P 100 , to obtain the
•• In calculus, we call 100 a critical number of P, because it is a
number in Dom P where P is either 0 in value or is undefined.
( )
We are interested in critical numbers, because those are the possible
x-coordinates of turning points on the graph of P.
WARNING 3: Some precalculus sources use the term critical
number differently; it is where P, itself, is either 0 in value or is
undefined. P values could change sign around critical numbers
and discontinuities.
(Section 1.11: Limits and Derivatives in Calculus) 1.11.15
• Find x-coordinates of points where the tangent lines slope upward;
their slopes are positive.
Solve:
()
P x > 0
2x + 200 > 0
2x > 200
x < 100
(
)
Therefore, P is increasing on the x-interval 0, 100 . In fact, we can
say that P is increasing on 0, 100 by a continuity argument.
(See Section 1.2, Part H on when a function increases on an interval.)
• Find x-coordinates of points where the tangent lines slope downward;
their slopes are negative.
Solve:
()
P x < 0
2x + 200 < 0
2x < 200
x > 100
(
)
Therefore, P is decreasing on the x-interval 100, . In fact, we can
)
say that P is decreasing on 100, by a continuity argument.
Here is the graph of P; the horizontal tangent line is in red:
§
(Section 1.11: Limits and Derivatives in Calculus) 1.11.16
FOOTNOTES
1. Etymology. The word “secant” comes from the Latin “secare” (to cut). The word “tangent”
comes from the Latin “tangere” (to touch). A secant line to the graph of f must intersect it in
at least two points. A tangent line only need intersect the graph in one point; the tangent line
might “just touch” the graph at that point (though there could be other intersection points).
2. Difference quotients with negative denominators. Our forms of difference quotients allow
negative denominators, as well. They still represent slopes of secant lines.
Form 1
(
)
• Form 1 b < a
slope =
()
()
() ()
( )
()
()
()
()
() ()
( )
()
()
f b f a f b f a
rise f a f b
=
= =
run
ab
b a
b a
(
• Form 2 x < a
slope =
Form 2
)
f x f a f x f a
rise f a f x
=
= =
run
ax
xa
xa
Form 3
(
• Form 3 h < 0
slope =
(
()
(
)
(
)
()
(
)
()
)
(
)
()
(
)
()
f a + h f a f a + h f a
rise f a f a + h
=
= =
run
h
h
a a+h
• Form 4 h < 0
slope =
)
Form 4
)
()
(
(
)
f x + h f x f x + h f x
rise f x f x + h
=
= =
run
h
h
x x+h
(
)