Understanding Composite Functions & Instantaneous Rates of Change in Math, Study notes of Calculus

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MATH 150 – TOPIC 3

FUNCTIONS: COMPOSITION AND DIFFERENCE QUOTIENTS

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

2 x − 1, 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 f (x) =

x and g(x) = 2x−1 (think of f ( ) =

( ) 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: f (g(x)) = f ( ) =

Now fill in the blanks with the g(x) representation and you get f (g(x)) = f (2x − 1) =

2 x − 1.

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 f (x) =

x, we have g(f (x)) = g(

x) = 2(

x) − 1 = 2

x − 1.

Alternative notation for composite functions: g(f (x)) = g ◦ f, f (g(x)) = f ◦ g.

Example: Let g(x) = x^2 + 1. Evaluate and simplify the difference

quotient

g(x) − g(2) x − 2 , x 6 = 2.

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

g(x) − g(2) x − 2

(x^2 + 1) − 5 x − 2

x^2 − 4 x − 2 = x + 2.

Example: Let g(x) =

x − 2

. Evaluate and simplify the difference

quotient g(x + ∆x) − g(x) ∆x

, ∆x 6 = 0 [∆x represents the “change” in x.]

Remember: g( ) =

g(x + ∆x) − g(x) ∆x

(x + ∆x) − 2

x − 2 ∆x = (x − 2) − (x + ∆x − 2) (x + ∆x − 2)(x − 2)

∆x

=

−∆x (x + ∆x − 2)(x − 2)(∆x)

=

(x + ∆x − 2)(x − 2)

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.

f (x)

• • (x + h, f (x + h))

(x, f (x))

x x^ +^ h h

m =

y 2 − y 1 x − 2 − x 1

∆y ∆x

f (x + h) − f (x) (x + h) − x

f (x + h) − f (x) h

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.

x

y = f (x)

tangent line•^ • ↘

h h 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

Math 150 T3-Functions and Difference Quotients Review – Answers Page 7

ANSWERS to PRACTICE PROBLEMS (Topic 3 – Functions and Difference Quotients)

3.1 a) i)

x

x

x^2

x

ii)

x^2 − 2 x

iii) (x^2 − 2 x)^2 − 2(x^2 − 2 x) = (x^2 − 2 x)(x^2 − 2 x − 2)

b) i) sin

x

ii)

sin x

c) inner: h(x) = 3x middle: g(x) = sin x outer: f (x) =

x CHECK: f (g(h))(x) = f (g(3x)) = f (sin 3x) =

sin 3x

h

g

f

Return to Problem

3.2 a) m = 3 b) m is undefined c) m = f (x + h) − f (x) h

Return to Problem

Math 150 T3-Functions and Difference Quotients Review – Answers Page 8

3.3 a)

(x + h)^2 − 3(x + h) − (x^2 − 3 x) h

=

2 xh + h^2 − 3 x h = 2x + h − 3

b)

x + h − 2 −

x − 2 h

x + h − 2 +

x − 2 √ x + h − 2 +

x − 2

=

x + h − 2)^2 − (

x − 2)^2 h(

x + h − 2 +

x − 2)

=

x + h − 2 − (x − 2) h(

x + h − 2 +

x − 2)

=

x + h − 2 +

x − 2

Return to Problem

2 + (x + ∆x)

2 + x ∆x

=

2 + x − (2 + x + ∆x) (2 + x + ∆x)(2 + x)∆x

=

(2 + x + ∆x)(2 + x)

Return to Problem

Beginning of Topic 150 Review Topics Skills Assessment