Lesson 4
Difference Quotients
August 26, 2013
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Quiz 1
(Average: 8.6/10)
1. Solve
3x 4 + x 2 = 2x 4 + 10x 2 .
Answer: Subtract the right-hand side to obtain x 4 − 9x 2 = 0. This
factors as x 2 (x − 3)(x + 3) = 0. Therefore x = −3, 0, 3.
2. Solve
x 2 + 3x + 2
= 0.
x 2 + 4x + 3
Answer: Factor the numerator and denominator.
(x + 1)(x + 2)
= 0.
(x + 1)(x + 3)
The potential answers are −1 and −2, but x + 1 appears in the
denominator, so the only answer is x = −2.
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Homework 3 Questions?
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Previously...
(Section 1.1)
1. What are functions?
2. Composing two functions.
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Overview
Today we finish Section 1.1 by discussing difference quotients.
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What’s a difference quotient?
Sometimes, when we talk about functions, we are interested in the average
rate of change of that function.
Lesson 4 Difference Quotients
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What’s a difference quotient?
Sometimes, when we talk about functions, we are interested in the average
rate of change of that function.
Let f (x) be a function. Then the average rate of change of f (x) between
x and x + h is given by the difference quotient:
f (x + h) − f (x)
,
h
or perhaps more usefully:
1
[f (x + h) − f (x)].
h
Lesson 4 Difference Quotients
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What’s a difference quotient?
Sometimes, when we talk about functions, we are interested in the average
rate of change of that function.
Let f (x) be a function. Then the average rate of change of f (x) between
x and x + h is given by the difference quotient:
f (x + h) − f (x)
,
h
or perhaps more usefully:
1
[f (x + h) − f (x)].
h
Later, we will use difference quotients in order to define derivatives.
Lesson 4 Difference Quotients
6 / 16
What’s a difference quotient?
Sometimes, when we talk about functions, we are interested in the average
rate of change of that function.
Let f (x) be a function. Then the average rate of change of f (x) between
x and x + h is given by the difference quotient:
f (x + h) − f (x)
,
h
or perhaps more usefully:
1
[f (x + h) − f (x)].
h
Later, we will use difference quotients in order to define derivatives.
All four problems in Lesson 4 involve calculating difference quotients.
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Example 1
Find
f (x + h) − f (x)
when f (x) = x + 1.
h
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Example 1 (Worked)
Find
f (x + h) − f (x)
when f (x) = x + 1.
h
1
1
[f (x + h) − f (x)] = [(x + h + 1) − (x + 1)]
h
h
1
= [h]
h
= 1.
Warning: The usual care we need to take with parentheses is doubled
when dealing with a difference quotient. That’s why some people might
prefer using h1 [f (x + h) − f (x)].
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Example 2
Find
f (x + h) − f (x)
when f (x) = x 2 .
h
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Example 2 (Worked)
Find
f (x + h) − f (x)
when f (x) = x 2 .
h
f (x + h) − f (x)
1
= [(x + h)2 − x 2 ]
h
h
1
= [x 2 + 2xh + h2 − x 2 ]
h
1
= [2xh + h2 ]
h
= 2x + h.
Warning: Notice that f (x + h) here is (x + h)2 and not x + h2 . Be
careful with parentheses!
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Example 3
Find
f (x + h) − f (x)
1
when f (x) = .
h
x
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Example 3 (Worked)
Find
f (x + h) − f (x)
1
when f (x) = .
h
x
1
1
1
f (x + h) − f (x)
=
−
h
h x +h x
1 x − (x + h)
=
h
x(x + h)
1
−h
=
h x 2 + xh
−1
= 2
.
x + xh
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Example 4
Find
f (x + h) − f (x)
1
when f (x) = 2 .
h
x
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Example 4 (Worked)
Find
f (x + h) − f (x)
1
when f (x) = 2 .
h
x
1
1
f (x + h) − f (x)
1
=
−
h
h (x + h)2 x 2
1 x 2 − (x + h)2
=
h
x 2 (x + h)2
2
1 x − x 2 − 2xh − h2
=
h
x 2 (x + h)2
−2x − h
.
= 4
x + 2x 3 h + x 2 h2
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Review
For Wednesday:
1. Attack homework 4!
2. Get ready for lots of word problems on Wednesday and Friday: find
the formulas for:
I
Area of a rectangle,
I
Perimeter of a rectangle,
I
Volume of a box,
I
Surface area of a box,
I
Volume of a cylinder.
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Quiz 2
Let f (x) = 6x 2 + 3 and g (x) =
1
.
x +1
1. Find f (g (x)) and simplify.
2. Find g (f (x)) and simplify.
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