This document discusses key concepts related to rates of change and derivatives: 1) It defines average rate of change (ARC) as the slope of a secant line on a graph or using the slope formula algebraically, and instantaneous rate of change (IRC) as the slope of the tangent line. 2) It introduces the difference quotient as a way to define ARC and IRC algebraically without a graph by taking the limit as h approaches 0. 3) A derivative is defined as a function that gives the IRC, allowing it to be evaluated at any point without graphing by taking the limit of the difference quotient.

1. 7.3 and part of 7.5
Rates of Change, Slopes, and Derivatives
A. What algebra skill you’ll need
B. Where average rate of change (ARC) is on a graph
C. How to find it algebraically (without a graph)
D. How we get the difference quotients definition of ARC
E. Where instantaneous rate of change (IRC) is on a graph
F. How we get the difference quotient definition of IRC
G. How to find the IRC
H. What a derivative is
I. What Leibniz’s Notation looks like
J. Word Problems

2. A. What algebra skills you’ll need
• Negative, positive slope
• Slope formula
• Combining rational expressions

3. B. Where average rate of change (ARC) is on
a graph The average rate of
change(ARC)
between x = 2
and x = 5 is the
SLOPE of that red
dotted line.
[a secant line]
Let’s find it:

4. C. How to find it algebraically (without a
graph)
• Well, since the ARC is really a SLOPE, let’s recall the
slope formula:
rise y2 − y1 f ( x2 ) − f ( x1 )
= =
run x2 − x1 x2 − x1
• We can use that f(x) notation because we’ll be
dealing with functions. Recall that f(x) simply
means the y value that corresponds to x.
• Let’s apply this to a problem:

5. " If f ( x) = 3 x + 5, find the average
2
rate of change between x = 1 and x = 2."

6. You try : " If f ( x) = x + 1, find the average
2
rate of change between x = 3 and x = 5."

7. D. How we get the difference quotients
definition of ARC
• You may or may not have been introduced to the
famous “difference quotient” in your algebra class.
Here is what it looked like:
f ( x + h) − f ( x)
h
• First, I’ll show you where this came from.
• Then , we’ll discover what is so great about it.

8. Let’s take this curve, and label a fixed point x. It
would have corresponding y-value labeled f(x).
If I jog over a
certain
distance h on
the x-axis.
What could I
call this new
fixed point?

9. x + h would be that newly created point
on the x-axis. Its corresponding
y-value would be called f(x + h).

10. Here is the red, dotted secant line.
We need the slope of it.

11. RISE =
RUN =
That’s where the difference quotient came from.
Now, I’ll show you what is so great about it.

12. E. Where instantaneous rate of change (IRC)
is on a graph
• It’s the slope of the tangent line. I’ll draw it:

13. F. How we get the difference quotient definition of IRC
Remember that secant line that was h wide?
• It was h
wide.
Imagine you
could make h
shrink
(“approach
zero”).

14. That would get closer and closer to the slope
of the tangent line! The IRC!
f ( x + h) − f ( x)
IRC = lim
h →0 h
This will be useful because we can use this
formula algebraically (no graphing necessary
to find or guess at the IRC.)
Let’s see if we can work one…

16. G. How to find the IRC
• Find the instantaneous rate of change of this
function at x = 1. f ( x) = x
2

17. 1
Find the slope of the tangent line to f ( x) = at x = 2.
x

18. 2
You try : Find the slope of the tangent line to f ( x) = at x = 3.
x

19. You try : Find the instantaneous rate
of change of f ( x) = 2 x at x = 3.
2

20. Sometimes they ask you whether the slope of
the tangent line is positive, negative, or zero.

21. H. What a derivative is
• We can evaluate the IRC at any point we want
algebraically, but wouldn’t it be easier if we
could create a function for the IRC, and then
we could plug in any value. It would be more
efficient.
• The DERIVATIVE is just that. Basically, it is a
function for the IRC.
• When you see “IRC,” think “Derivative.”

22. The derivative, noted by f ′( x ) , can be found by
f ( x + h) − f ( x)
f ′( x ) = lim .
h →0 h
Find the derivative of f ( x) = x 2 − 7 x + 150.

23. f ( x + h) − f ( x)
You try : f ′( x ) = lim
h →0 h
Find the derivative of f ( x) = x 2 − 2 x + 41.

24. I. What Leibniz’s Notation looks
like
d
Instead of writing f ′( x ) , Leibniz wrote f ( x ).
dx
dy
Instead of writing y′, Leibniz wrote .
dx

25. J. Word Problems
• First of all, the units of the ARC or IRC is
always consistent with it being a rate. [Like,
“something” per “something”]
• More particularly, the first “something” is the
units of the function f(x), and the second
“something” is the units of the x.
• “something” per “something”

26. Refining crude oil requires heating and cooling at different
rates. Suppose the temperature of the oil at time x hours is
f ( x) = x − 7 x + 150 degrees Fahrenheit. Find the IRC of
2
temperature at time x = 6, and interpret it.

27. You try : Suppose the temperature of the oil at time x hours is
f ( x) = x − 2 x + 12 degrees Fahrenheit. Find the IRC of
2
temperature at time x = 2, and interpret it.

28. Use the limit definition to find an EQUATION
(y=mx+b) of the tangent line to the graph of f at
the given point:
f ( x ) = − x ; ( − 1,− 1)
2

29. The word “differentiable” means…
_________________________________

30. • If it is differentiable at a point, then it is
continuous at that point.
• BUT
• Just because it is continuous at a point doesn’t
necessarily mean it is differentiable there.
• Example: