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Functions and Difference Quotients Review

I. Composition of Functions

II. Difference Quotients
Practice Problems

I. Composition of Functions

HAPPY BIRTHDAY... OK so in all likelihood today isn't your birthday.
Let's assume it is and you've been given a present wrapped in a large box.
You open the box and surprise!!!, inside is another box. Undaunted you
open it and discover yet another box. This may appear to be a lesson in ac-
cepting disappointment. Unbeknownst to you, though, you are experiencing
composite function behavior.

HUH?!?! Let us explain.

The composite function is like having one function contained inside another.
When you see you probably think of it as just another function,
but it's something more. It's a composite (one function inside another).
Here's why.

Suppose and g(x) = 2x−1 (think of as the larger
box). Now place 2x − 1 inside. Mathematically this is written as f(g(x))
and we call it "f composite g". To form a composite, try the following:

Now ll in the blanks with the g(x) representation and you get

NOTE: just like f(3) means to input 3 into f, f(g(x)) means to input g
into f.

Unlike boxes, any function can be placed inside another. From above,
g(x) = 2x −1. This really means g( )=2( )− 1. If . we have

Alternative notation for composite functions:

Let's try a different practice drill. Suppose h(x) = sin(2x). Can you state
functions for f and g so that h = f(g(x))?

Solution: inner function is 2x, which is g,
outer function is sin x = sin( ), which is f.

Check: f(g(x)) = f( )=sin( ).
Now insert 2x and get f(g(x)) = f(2x) = sin2x.

NOTE: This is the most obvious choice for f and g but not the only
one. Let us try f(x) = sin(x + 1) and g(x) = 2x − 1.

Check: f(g(x)) = f( ) = sin[( ) + 1],
so f(g(x)) = f(2x − 1) = sin[(2x − 1) + 1] = sin 2x.

What's so special about composite functions? One day in Calc I your
instructor will throw a birthday party for the entire class, and his present
to you will be called the CHAIN RULE.

Exercise: Practice Problem 3.1.

II. Difference Quotients

A. Introduction

In algebra, rate of change is introduced in its most basic form by finding
the slope of a line using . This "formula" can also be called
a Difference Quotient. In calculus, rates of change (both average and
instantaneous) are found using function forms of difference quotients.

Let's start with a few examples of the algebra in difference quotients.

Example: Let g(x) = x2 + 1. Evaluate and simplify the difference

Solution: Using g(x) = x2 + 1 and g(2) = 5,

Example: Let Evaluate and simplify the difference
quotient [ Δx represents the "change" in x.]


B. Application

How does a difference quotient measure rate of change (or slope)?

Suppose we start with a function f(x) and draw a line (called a secant
line) connecting two of its points.

Notice that the slope is represented by a difference quotient and is
referred to as the average rate of change. Now suppose the distance
between these two points is shortened by decreasing the size of h.

Conclusion: The two points get so close together they almost con-
cide. The resulting line is now a tangent line whose slope measures the
instantaneous rate of change. You will come to know this slope by the
term "derivative".

PRACTICE PROBLEMS for Topic 3 - Functions and Difference Quotients

3.1 a) Suppose f(x) = x2 − 2x and Form the following

b) Suppose f(x) = sinx and Form the following compositions:

c) Suppose Select functions for f, g, and h so that
y = f(g(h))(x). Check your results.

Return to Review Topic Answers

3.2 Find the rate of change (slope) of the line containing the given points.

a) (2,−5) and (4,1)

b) (3,−4) and (3,−1)

c) (x, f(x)) and (x + h, f(x + h))


3.3 Evaluate and simplify the difference quotient when:

a) f(x) = x2 − 3x,

b) (Rationalize the numerator).


3.4 Evaluate and simplify the difference quotient


(Topic 3 - Functions and Difference Quotients)

c) inner: h(x) = 3x
middle: g(x) = sinx

Return to Problem

a) m = 3

b) m is undefined

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