Section Composite and Inverse Functions

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Section 8.4 - Composite and Inverse Functions

I. Composition of Functions

A. If f and g are functions, then the composite function of f and g (written f



g) is:

(f



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

The domain of f



g is the set of all x in the domain of g such that g(x) is in the domain of f.

B. With composition, we are, in effect, substituting a number into g(x), finding out what y is, and then substituting that answer into f(x).

C. Examples - Let f(x) = 9 − 2x, g(x) = −5x + 2. find the following.

1. (f



g)(x)

First, we use the definition of composition to get:

(f



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

Now we will substitute into this equation what g(x) is equal to:

(f



g)(x) = f(−5x + 2)

Next, we substitute −5x + 2 in for x in f(x), EVEN THOUGH x IS REPEATED!

(f



g)(x) = 9 − 2(−5x + 2) Simplifying, we get:

Answer: (f



g)(x) = 5 + 10x 2. Now you try one: (g



f)(x)

Answer: (g



f)(x) = −−−−43 + 10x

Note that composition, in general, is not commutative.

3. (f



g)(3)

Again, we start by using the definition of composition to get:

(f



g)(3) = f(g(3))

Substituting 3 for x in g(x), we get:

(f



g)(3) = f(−5(3) + 2) = f(−13)

We now substitute −13 in for x in f(x) to get:

(f



g)(3) = 9 − 2(−13) Simplifying, we get:

Answer: (f



g)(3) = 35 4. Now you try one: (g



f)(3)

Answer: (g



f)(3) = −−−−13

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5.

a. (f



g)( −2)

We start by using the definition of composition:

(f



g)( −2) = f(g(−2))

We now have to determine the value of y when x is −2 for the graph of g(x):

(f



g)( −2) = f(2)

We next look at the graph of f(x) and determine the value of y when x is 2:

Answer: (f



g)( −2) = 3 b. Now you try one: (g



f)( −4)

Answer: (g



f)( −4) = 2 II. Inverse Properties

A. Recall that for a real number A, the additive inverse was that real number B such that A + B = 0.

B. For a real number A ≠ 0, the multiplicative inverse is that real number B such that AB = 1.

C. For a function f(x), the inverse function is that function g(x) such that

(

f g

)

(x) = x and

(

g  f

)

(x) = x.

D. Verifying that functions are inverses of each other.

1. Do the composition

(

f g

)

(x). If the answer is x, you are halfway there.

2. Now do the composition

(

g  f

)

(x). If this answer is also x, then f and g are inverse functions of each other. We then would write that g(x) = f -1 (x). The "-1" is NOT an exponent. This notation means that we have the inverse function of f(x). Note that f(x) is also g -1 (x).

E. Examples - Determine if f(x) and g(x) are inverses of each other.

1. f(x) = 3x−4, g(x) = x3 + 4 We first do

(

f g

)

(x).

(

f g

)

(x) = f(g(x)) = f(x3 + 4) Now substitute this in for x in f.

= 3(x3+4)− =4 3x3+ − =4 4 3x3 =x So this is half right. Now we do

(

g  f

)

(x).

(

g  f

)

(x) = g(f(x)) = g(3x−4) = (3x−4)3 + 4 = x - 4 + 4 = x Answer: f(x) and g(x) are inverses.

y = f(x) y = g(x)

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2. Now you try one: f(x) = 5x − 9, g(x) = 5 9 x+

Answer: f(x) and g(x) are not inverses.

III. Inverse Functions

A. For a function f(x), the inverse function is that function g(x) such that

(

f g

)

(x) = x and

(

g  f

)

(x) = x.

B. Verifying that functions are inverses of each other.

1. Do the composition

(

f g

)

(x). If the answer is x, you are halfway there.

2. Now do the composition

(

g  f

)

(x). If this answer is also x, then f and g are inverse functions of each other. We then would write that g(x) = f1 (x). The "−1" is NOT an exponent. This notation means that we have the inverse function of f(x). Note that f(x) is also g1 (x).

IV. Determining if a Function has an Inverse

A. A function f(x) is one-to-one if for every y in the range there is only one x in the domain that corresponds to it.

B. Horizontal Line Test: A function f(x) is not one-to-one if any horizontal line intersects the graph of f(x) in more than one point.

C. If a function f(x) is one-to-one, then its inverse is also a function. When this occurs, we write the inverse function as f -1 (x) (read "f inverse of x"). Note that this is the functional inverse, NOT the multiplicative inverse.

D. Examples - Are these functions one-to-one?

1.

Answer: Not one-to-one.

2.

Answer: Yes one-to-one.

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3. Now you try one:

V. Finding the Inverse

A. In general, to find the inverse of a relation, we switch x & y in the ordered pairs. Remember that x is the domain, y is the range.

B. This means, geometrically, that the graph of a relation and its inverse are reflections of each other across the identity function line, f(x) = x.

C. Finding the inverse of a function f(x)

1. Determine if the function is one-to-one.

2. Write y for f(x).

3. Switch x & y.

4. Solve for y.

5. Write f 1 (x) for y.

6. Verify by showing that

(

f  f1

) (

(x)= f1 f

)

(x)= x.

7. Remember:

a. Domain of f is the range of f 1. b. Range of f is the domain of f 1. D. Examples - Find the inverse function.

1. f(x) = 4x − 5

First, we write y for f(x).

y = 4x − 5

Next, switch x & y.

x = 4y − 5

Now we solve for y.

x + 5 = 4y OR x y

+ = 4 5

Answer: f −−−−1 (x) = 4 x 5 4 1 + Verify:

(

f  f1

)

(x)= f(f 1 (x)) = f14x+45 = 414x+45−5 = x + 5 − 5 = x So this is ok.

You verify that

(

f1 f

)

(x)=x.

Note that if we graph these, we make a table for the function that is the "easiest", then switch x & y to get the table for the inverse.

Answer: Yes one-to-one.

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2. f(x) = 3x−5

First, we replace f(x) with y.

y = 3x−5

Now we switch x & y.

x = 3y−5

Now we solve for y.

x3 = y − 5 OR x3 + 5 = y Answer: f −−−−1 (x) = x3 + 5

Verify:

(

f  f1

)

(x)= f(f 1 (x)) = f(x3 + 5) = 3

( )

x3+5 5 = 3 3x = x. So this is ok.

You verify that

(

f1f

)

(x)= x.

3. Now you try one: f(x) = x3 + 1 Answer: f −−−−1 (x) = 3x1 VI. Graphing a function and its inverse.

A. Remember that to find the inverse, we switch x & y.

B. So if we make a table for f(x), to get a table for f −1 (x), we just switch x & y on the table.

C. Examples - Graph f(x) and f1 (x) on the same set of axes.

1. f(x) = 4x − 5, f 1 (x) =1 5 4x +4

Making a table for f(x) will be relatively easy, but f1 (x) doesn't look so nice!

Now graph both of these.

x f(x) = 4x −−−− 5

0 −5

1 −1

x f −−−−1 (x) =1 5 4x +4

−5 0

−1 1

Switch x & y on the table to get the table for f1 (x).

f(x)

f−−−−1 (x)

Figure

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