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SAT

 


 SAT Math Lesson Plan 20: Function Composition and Inverses


 

Lesson Summary

This lesson covers the concepts of function composition and inverses, which are critical topics for the SAT Math section. Approximately 7-10% of SAT Math questions involve function composition and inverse functions, requiring a strong understanding of how to combine functions, find inverses, and verify inverse relationships algebraically and graphically. The lesson introduces these concepts step by step, helping students develop the skills necessary for identifying and solving related problems.

Lesson Objectives

  • Perform function composition and interpret its meaning.
  • Find the inverse of a function algebraically.
  • Verify inverses by applying function composition.
  • Analyze inverse functions graphically.

Common Core Standards

  • CCSS.MATH.CONTENT.HSF.BF.A.1.C – Compose functions to model situations.
  • CCSS.MATH.CONTENT.HSF.BF.B.4.A – Determine the inverse of a function.
  • CCSS.MATH.CONTENT.HSF.IF.B.4 – Interpret functions that arise in real-world applications.

Prerequisite Skills

  • Understanding of basic function notation.
  • Ability to evaluate functions for given inputs.
  • Knowledge of solving equations algebraically.

Key Vocabulary

  • Function Composition: Applying one function to the output of another: (f ∘ g)(x) = f(g(x)).
  • Inverse Function: A function that reverses the effect of another function.
  • One-to-One Function: A function where each input corresponds to a unique output, ensuring an inverse exists.
  • Horizontal Line Test: A test to determine if a function has an inverse by checking if any horizontal line crosses the graph more than once.
  • Reflection Over y = x: The graphical representation of inverse functions.

 


 

Warm-Up

Activity 1: Review of Inverse Operations

In this activity, we’ll review how inverse operations work. Inverse operations are pairs of operations that "undo" each other. Let’s look at the following equations:

 

Operation

Inverse

Addition

Subtraction

Multiplication

Division

 

Let’s apply these to some simple equations:

  1. \( x + 5 = 12 \) (Subtract 5 from both sides: \( x = 7 \))
  2. \( 4x = 20 \) (Divide both sides by 4: \( x = 5 \))

The idea of inverse operations is important to think about as we think about inverse functions.

Activity 2: Review of Reflections

In this activity, we will review geometric reflections. A reflection is a transformation where each point of a figure is mapped to a corresponding point on the opposite side of a line, such as a line of symmetry.

Consider the following example:

 

reflections

 

In this example, the points are reflected across the y-axis. You can see how each point is mapped to the opposite side of the line of reflection. Similarly, we can reflect a graph of a function across a vertical line such as the line \( x = 0 \). This is a foundational concept in understanding transformations of functions.

Activity 3: Review of Function Notation

In this activity, we’ll review various types of function notation and their interpretations. Understanding how to work with function notation is critical for solving SAT Math problems. Here’s a summary of different expressions and their meanings:

 

ExpressionExplanationExample
\( f(x) \)Represents a function \( f \) applied to an input \( x \). This is the basic form of function notation.\( f(x) = 2x + 3 \), so \( f(2) = 2(2) + 3 = 7 \)
\( f(x + k) \)Represents a horizontal shift of the graph of \( f(x) \) by \( k \) units. If \( k \) is positive, the shift is to the left; if \( k \) is negative, the shift is to the right.\( f(x + 3) = 2(x + 3) + 3 = 2x + 9 \)
\( f(x) + g(x) \)Represents the sum of two functions \( f(x) \) and \( g(x) \). The sum function is created by adding the individual outputs of the two functions for each input \( x \).\( f(x) = 2x + 3 \), \( g(x) = x - 1 \), so \( f(x) + g(x) = (2x + 3) + (x - 1) = 3x + 2 \)
\( f(x) \cdot g(x) \)Represents the product of two functions \( f(x) \) and \( g(x) \). The product function is created by multiplying the individual outputs of the two functions for each input \( x \).\( f(x) = 2x + 3 \), \( g(x) = x - 1 \), so \( f(x) \cdot g(x) = (2x + 3)(x - 1) = 2x^2 + x - 3 \)

 

This table summarizes some of the most common forms of function expressions you will encounter. Understanding these forms and how they relate to the graphs of functions will help you solve more complex SAT problems.

 


 

Teach

In this section, we’ll explore the core concepts of functions, including how they work, how to compose them, and how to manipulate them algebraically. These concepts are essential for understanding SAT Math problems and will lay the foundation for more complex mathematical operations. Here’s what we will cover:

  • Function Notation: A review of how functions are written and how to evaluate them.
  • Function Composition: How to combine two functions to form a new one and solve for the resulting expression.
  • Inverse Functions: Understanding and finding the inverse of a function, and how it relates to the original function.
  • Transformations of Functions: Identifying how shifts, stretches, and reflections change the graph of a function.
  • Asymptotes: Understanding the behavior of rational functions and how to identify vertical and horizontal asymptotes.

Now, let’s dive deeper into each of these concepts and apply them through examples. Understanding function composition, as well as working with inverses, is vital in recognizing patterns and solving problems efficiently on the SAT.

Example 1: Function Composition

Let’s consider the two functions \( f(x) = 2x + 3 \) and \( g(x) = x - 1 \).

The composition \( (f \circ g)(x) \) means that we substitute \( g(x) \) into \( f(x) \). Here’s how we do it:

 

StepExplanation
1Substitute \( g(x) = x - 1 \) into \( f(x) \). This gives: \( f(g(x)) = 2(x - 1) + 3 \).
2Simplify: \( f(g(x)) = 2x - 2 + 3 = 2x + 1 \).

 

Thus, the composition \( (f \circ g)(x) = 2x + 1 \).

Example 2: Finding the Inverse of a Function

Now, let’s find the inverse of the function \( f(x) = 2x + 3 \).

To find the inverse, follow these steps:

  1. Start with the function: \( y = 2x + 3 \).
  2. Swap \( x \) and \( y \): \( x = 2y + 3 \).
  3. Solve for \( y \): \( x - 3 = 2y \), then \( y = \frac{x - 3}{2} \).

The inverse function is \( f^{-1}(x) = \frac{x - 3}{2} \).

To verify that \( f(x) \) and \( f^{-1}(x) \) are indeed inverses, we check if:

  • \( (f \circ f^{-1})(x) = x \)
  • \( (f^{-1} \circ f)(x) = x \)

Let’s check:

 

StepExplanation
1Substitute \( f^{-1}(x) = \frac{x - 3}{2} \) into \( f(x) = 2x + 3 \).
2Simplify: \( f(f^{-1}(x)) = 2\left(\frac{x - 3}{2}\right) + 3 = x - 3 + 3 = x \).

 

Since \( (f \circ f^{-1})(x) = x \), we have confirmed that \( f(x) \) and \( f^{-1}(x) \) are inverses.

Example 3: Finding the Inverse of a Quadratic Function

In this example, we will find the inverse of a quadratic function. Since quadratic functions are not one-to-one, we must restrict the domain to ensure that the inverse function exists.

Given the quadratic function:

\[ f(x) = x^2 - 4x + 3 \]

We will now find its inverse.

Step 1: Swap \( f(x) \) and \( y \)

To find the inverse, start by replacing \( f(x) \) with \( y \), so:

\[ y = x^2 - 4x + 3 \]

Step 2: Solve for \( x \) in terms of \( y \)

Now, swap \( x \) and \( y \), and solve for \( y \). We begin by completing the square to make the equation easier to manipulate:

StepExplanation
1Start with the equation: \( x = y^2 - 4y + 3 \)
2Complete the square: \( x = (y - 2)^2 - 1 \)
3Solve for \( y \): \( y = \pm \sqrt{x + 1} + 2 \)

Thus, the inverse is:

\[ f^{-1}(x) = \pm \sqrt{x + 1} + 2 \]

Since the quadratic is not one-to-one, we restrict the domain of the inverse. For the function to be invertible, we choose either the positive or negative square root to ensure the inverse is a function.

Step 3: Verify the inverse using composition

To verify the inverse, we perform the following compositions:

\[ f(f^{-1}(x)) = x \] \[ f^{-1}(f(x)) = x \]

Both compositions should simplify back to \( x \), confirming that the inverse we found is correct. You can check this using a graph.

Step 4: Graphing the Function and its Inverse

Here is the graph of the function \( f(x) = x^2 - 4x + 3 \) and its inverse. Notice that the inverse is restricted to either the positive or negative branch of the parabola.

 

FunctionsFunctions

 

Desmos Link: Use Desmos to graph the function and the inverse, and explore the domain/range restrictions.

Example 4: Identifying the Inverse of a Function

Consider the function \( f(x) = 3x - 2 \). Let’s find its inverse.

Follow these steps:

  1. Start with the function: \( y = 3x - 2 \).
  2. Swap \( x \) and \( y \): \( x = 3y - 2 \).
  3. Solve for \( y \): \( x + 2 = 3y \), then \( y = \frac{x + 2}{3} \).

The inverse function is \( f^{-1}(x) = \frac{x + 2}{3} \).

Example 5: Verifying the Inverse Using Composition

We will show that \( f(x) = 3x + 1 \) and its inverse \( f^{-1}(x) = \frac{x - 1}{3} \) are true inverses of each other by performing both compositions.

Step 1: Find the inverse of \( f(x) = 3x + 1 \)

To find the inverse, swap \( f(x) \) and \( x \), and then solve for \( y \):

\[ f(x) = 3x + 1 \quad \Rightarrow \quad y = 3x + 1 \] Now swap \( x \) and \( y \), and solve for \( y \): \[ x = 3y + 1 \] \[ x - 1 = 3y \] \[ y = \frac{x - 1}{3} \] Thus, the inverse is \( f^{-1}(x) = \frac{x - 1}{3} \).

Step 2: Verify the inverse using composition

We will now verify the inverse by performing both compositions:

Composition 1: \( f(f^{-1}(x)) \)

Substitute \( f^{-1}(x) = \frac{x - 1}{3} \) into the original function \( f(x) = 3x + 1 \):

\[ f(f^{-1}(x)) = 3\left(\frac{x - 1}{3}\right) + 1 \] \[ = x - 1 + 1 \] \[ = x \]

Composition 2: \( f^{-1}(f(x)) \)

Now, substitute \( f(x) = 3x + 1 \) into the inverse \( f^{-1}(x) = \frac{x - 1}{3} \):

\[ f^{-1}(f(x)) = \frac{(3x + 1) - 1}{3} \] \[ = \frac{3x}{3} \] \[ = x \]

Verification Table:

 

CompositionResult
\( f(f^{-1}(x)) \)\( x \)
\( f^{-1}(f(x)) \)\( x \)

 

Both compositions confirm that \( f(x) \) and \( f^{-1}(x) \) are indeed inverses of each other, as the result is \( x \) in both cases.

Example 6: Graphing a Function and Its Inverse

Consider the function \( f(x) = 2x + 1 \) and its inverse \( f^{-1}(x) = \frac{x - 1}{2} \).

Let's plot both the function and its inverse on a graph to visually verify the inverse relationship.

 

Graph of \( f(x) = 2x + 1 \)

Graph of \( f^{-1}(x) = \frac{x - 1}{2} \)

FunctionsFunctions

 

Desmos Link: Use Desmos to graph these functions and explore their relationship.

Example 7: Real-World Example - Function Composition in a Word Problem

A car rental company charges a base fee of \$30 per day for renting a car. Additionally, they charge \$0.15 per mile driven. Let’s define the following two functions:

  • \( C(x) \) represents the cost of renting the car for \( x \) days.
  • \( D(y) \) represents the distance driven in \( y \) miles.

Find the total cost for a car rental where a customer rents a car for 4 days and drives 200 miles.

Solution:

Step 1: Write down the given functions:

  • \( C(x) = 30x \) (Cost of renting a car for \( x \) days).
  • \( D(y) = 0.15y \) (Cost per mile driven, for \( y \) miles).

Step 2: Substitute the appropriate values into the functions:

  • First, calculate the cost for 4 days: \( C(4) = 30 \times 4 = 120 \).
  • Next, calculate the cost for driving 200 miles: \( D(200) = 0.15 \times 200 = 30 \).

Step 3: Combine the results:

  • The total cost will be the sum of both functions: \( \text{Total Cost} = C(4) + D(200) = 120 + 30 = 150 \).

Why This Example Matters:

  • Concepts Tested: Function composition and understanding how different types of functions can be combined to form a final result.
  • Real-World Context: The problem presents a realistic scenario that students may encounter in the context of a car rental service, allowing them to apply their understanding of functions and composition to solve a problem.
  • Multiple Function Uses: The example tests the student’s ability to evaluate each function independently and then combine the results.

  


 

Review

In this review section, we will summarize the key concepts covered in this lesson. These include:

  • Understanding function notation and its importance in expressing mathematical relationships.
  • Performing function composition and understanding the resulting function.
  • Identifying and finding the inverse of a function algebraically.
  • Verifying inverse functions using function composition.
  • Analyzing rational functions and their asymptotes (vertical and horizontal).

Example 1: Function Composition

Given the functions \( f(x) = 2x + 1 \) and \( g(x) = x^2 - 3 \), find the composition \( (f \circ g)(x) \):

Start by substituting \( g(x) \) into \( f(x) \), so:

 

StepExplanation
1Substitute \( g(x) = x^2 - 3 \) into \( f(x) \): \( f(g(x)) = 2(x^2 - 3) + 1 \).
2Simplify: \( f(g(x)) = 2x^2 - 6 + 1 = 2x^2 - 5 \).

 

Thus, \( (f \circ g)(x) = 2x^2 - 5 \).

Example 2: Finding the Inverse of a Function

Given the function \( f(x) = 3x - 5 \), let’s find its inverse.

  1. Start with the function: \( y = 3x - 5 \).
  2. Swap \( x \) and \( y \): \( x = 3y - 5 \).
  3. Solve for \( y \): \( x + 5 = 3y \), then \( y = \frac{x + 5}{3} \).

The inverse function is \( f^{-1}(x) = \frac{x + 5}{3} \).

Example 3: Analyzing Asymptotes for Rational Functions

Given the rational function \( f(x) = \frac{1}{x - 2} \), determine its vertical and horizontal asymptotes:

  • Vertical Asymptote: Set the denominator equal to zero, \( x - 2 = 0 \), so \( x = 2 \).
  • Horizontal Asymptote: For rational functions where the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).

Thus, the vertical asymptote is at \( x = 2 \), and the horizontal asymptote is at \( y = 0 \).

Desmos Link: Use Desmos to graph the function and observe the asymptotic behavior.

Example 4: Function Composition with Linear Functions

Consider the functions \( f(x) = 5x - 3 \) and \( g(x) = 2x + 1 \). Find the composition \( (f \circ g)(x) \), which means substituting \( g(x) \) into \( f(x) \):

 

StepExplanation
1Substitute \( g(x) = 2x + 1 \) into \( f(x) \): \( f(g(x)) = 5(2x + 1) - 3 \).
2Simplify: \( f(g(x)) = 10x + 5 - 3 = 10x + 2 \).

 

Thus, \( (f \circ g)(x) = 10x + 2 \).

Desmos Link: Use Desmos to graph the function composition and explore its behavior.

Multimedia Resources

See a collection of Media4Math resources that support this topic. Click on this link.

 


 

Quiz

Directions: Solve each problem. Choose the correct answer from the options provided.

  1. What is the composition of \( f(x) = 2x + 3 \) and \( g(x) = x - 1 \), written as \( (f \circ g)(x) \)?
    a) \( 2x + 2 \)
    b) \( 2x + 1 \)
    c) \( 2x + 3 \)
    d) \( 2x + 4 \)

     
  2. What is the inverse of \( f(x) = 4x - 7 \)?
    a) \( f^{-1}(x) = \frac{x + 7}{4} \)
    b) \( f^{-1}(x) = \frac{x - 7}{4} \)
    c) \( f^{-1}(x) = \frac{7 - x}{4} \)
    d) \( f^{-1}(x) = \frac{x - 4}{7} \)

     
  3. Which of the following is the correct way to find the inverse of the function \( f(x) = 3x + 5 \)?
    a) Swap \( x \) and \( y \), then solve for \( y \)
    b) Add 5 to both sides and divide by 3
    c) Subtract 3 from both sides and divide by 5
    d) Add 5 to both sides and subtract 3

     
  4. What is the vertical asymptote of the function \( f(x) = \frac{1}{x - 4} \)?
    a) \( x = 4 \)
    b) \( y = 4 \)
    c) \( x = -4 \)
    d) \( y = -4 \)

     
  5. Which function represents the inverse of \( f(x) = 5x - 3 \)?
    a) \( f^{-1}(x) = \frac{x + 3}{5} \)
    b) \( f^{-1}(x) = \frac{x - 3}{5} \)
    c) \( f^{-1}(x) = \frac{x - 5}{3} \)
    d) \( f^{-1}(x) = \frac{x + 5}{3} \)

     
  6. What is the composition of \( f(x) = 2x + 1 \) and \( g(x) = x^2 - 3 \), written as \( (f \circ g)(x) \)?
    a) \( 2x^2 - 6 + 1 \)
    b) \( 2x^2 - 5 \)
    c) \( 2x^2 + 1 \)
    d) \( 2x^2 + 5 \)

     
  7. If the horizontal asymptote of a rational function is \( y = 0 \), what can we infer about the function?
    a) The function has a vertical asymptote at \( x = 0 \).
    b) The degree of the numerator is less than the degree of the denominator.
    c) The degree of the numerator is greater than the degree of the denominator.
    d) The function has no vertical asymptotes.

     
  8. Which of the following is the correct equation for finding the inverse of \( f(x) = 2x - 1 \)?
    a) \( y = \frac{x + 1}{2} \)
    b) \( y = \frac{x - 1}{2} \)
    c) \( y = \frac{1 - x}{2} \)
    d) \( y = \frac{x - 2}{1} \)

     
  9. What is the range of the function \( f(x) = 3x - 5 \)?
    a) All real numbers
    b) \( x > -5 \)
    c) \( x > 5 \)
    d) \( y > -5 \)

     
  10. Which of the following is the inverse of \( f(x) = 3x + 6 \)?
    a) \( f^{-1}(x) = \frac{x - 6}{3} \)
    b) \( f^{-1}(x) = \frac{x + 6}{3} \)
    c) \( f^{-1}(x) = \frac{x + 3}{6} \)
    d) \( f^{-1}(x) = \frac{x + 6}{9} \)

     

Answer Key

  1. Answer: b) \( 2x + 1 \)
    Explanation: To find the composition \( (f \circ g)(x) \), substitute \( g(x) = x - 1 \) into \( f(x) = 2x + 3 \). This gives \( f(g(x)) = 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1 \).

     
  2. Answer: b) \( f^{-1}(x) = \frac{x - 7}{4} \)
    Explanation: To find the inverse of \( f(x) = 4x - 7 \), swap \( x \) and \( y \), then solve for \( y \). \( x + 7 = 4y \), so \( y = \frac{x + 7}{4} \).

     
  3. Answer: a) Swap \( x \) and \( y \), then solve for \( y \).
    Explanation: To find the inverse, start with the equation \( y = 3x + 5 \), swap \( x \) and \( y \), then solve for \( y \). This gives \( y = \frac{x - 5}{3} \).

     
  4. Answer: a) \( x = 4 \)
    Explanation: The vertical asymptote is found by setting the denominator equal to zero. \( x - 4 = 0 \) gives \( x = 4 \).

     
  5. Answer: b) \( f^{-1}(x) = \frac{x - 3}{5} \)
    Explanation: To find the inverse of \( f(x) = 5x - 3 \), swap \( x \) and \( y \), then solve for \( y \). \( x + 3 = 5y \), so \( y = \frac{x + 3}{5} \).

     
  6. Answer: b) \( 2x^2 - 5 \)
    Explanation: To find \( (f \circ g)(x) \), substitute \( g(x) = x^2 - 3 \) into \( f(x) = 2x + 1 \). This gives \( f(g(x)) = 2(x^2 - 3) + 1 = 2x^2 - 6 + 1 = 2x^2 - 5 \).

     
  7. Answer: b) The degree of the numerator is less than the degree of the denominator.
    Explanation: For rational functions where the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at \( y = 0 \).

     
  8. Answer: b) \( y = \frac{x - 1}{2} \)
    Explanation: To find the inverse of \( f(x) = 2x - 1 \), swap \( x \) and \( y \), then solve for \( y \). This gives \( y = \frac{x + 1}{2} \).

     
  9. Answer: a) All real numbers
    Explanation: Since \( f(x) = 3x - 5 \) is a linear function, its range is all real numbers.

     
  10. Answer: a) \( f^{-1}(x) = \frac{x - 6}{3} \)
    Explanation: To find the inverse of \( f(x) = 3x + 6 \), swap \( x \) and \( y \), then solve for \( y \). This gives \( y = \frac{x - 6}{3} \).