2026年09月16日 Wednesday 14:02
🎯 Today's Objectives
- Students Will Be Able To add, subtract, multiply, and divide functions and state the domain of the resulting function.
- Students Will Be Able To form and evaluate composite functions using function notation.
- Students Will Be Able To determine the domain of a composite function and explain any restrictions.
- Students Will Be Able To apply operations and composition of functions to solve contextual problems.
🧰 Materials
- Whiteboard and markers
- Student handout with guided notes and practice problems
- Graphing calculators (optional, for checking)
- Exit ticket slips
- Diagnostic warm-up questions projected or printed
🖨️ Print student handout
⏱️ Suggested Time Allocation
| Segment |
Start |
End |
| Warm-Up |
0:00 |
0:05 |
| New Content + Examples |
0:05 |
0:22 |
| Guided Practice |
0:22 |
0:32 |
| Independent Practice |
0:32 |
0:45 |
| Exit Ticket |
0:45 |
0:50 |
🔥 Warm-Up
Given \(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\), evaluate \(f(4)\) and \(g(-2)\).
To evaluate \(f(4)\), substitute \(x = 4\) into \(f(x)\): \(f(4) = 2(4) + 3 = 8 + 3 = 11\). To evaluate \(g(-2)\), substitute \(x = -2\) into \(g(x)\): \(g(-2) = (-2)^2 - 1 = 4 - 1 = 3\).
Simplify \(\frac{x^2 - 9}{x - 3}\) and state any restrictions on \(x\).
Factor the numerator: \(x^2 - 9 = (x - 3)(x + 3)\). Then \(\frac{(x - 3)(x + 3)}{x - 3} = x + 3\), provided \(x \neq 3\) because the original denominator cannot be zero. So the simplified expression is \(x + 3\), \(x \neq 3\).
If \(h(x) = \sqrt{x}\), what is the domain of \(h(x)\)? Explain.
The domain of \(h(x) = \sqrt{x}\) is all real numbers \(x \geq 0\), because the square root of a negative number is not a real number. In interval notation, the domain is \([0, \infty)\).
📚 New Content
Key Concepts:
1. Operations on Functions: Given two functions \(f\) and \(g\), we can define new functions by performing arithmetic operations:
- Sum: \((f + g)(x) = f(x) + g(x)\)
- Difference: \((f - g)(x) = f(x) - g(x)\)
- Product: \((f \cdot g)(x) = f(x) \cdot g(x)\)
- Quotient: \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\), provided \(g(x) \neq 0\).
2. Domain of Combined Functions: The domain of \(f + g\), \(f - g\), and \(f \cdot g\) is the intersection of the domains of \(f\) and \(g\). For the quotient \(\frac{f}{g}\), the domain is the intersection of the domains of \(f\) and \(g\), excluding any \(x\) for which \(g(x) = 0\).
3. Composition of Functions: The composition of \(f\) and \(g\), denoted \(f \circ g\), is defined by \((f \circ g)(x) = f(g(x))\). The domain of \(f \circ g\) consists of all \(x\) in the domain of \(g\) such that \(g(x)\) is in the domain of \(f\).
4. Evaluating Composite Functions: To evaluate \((f \circ g)(a)\), first find \(g(a)\), then substitute that result into \(f\).
5. Important Observations:
- Function composition is not commutative: \(f \circ g \neq g \circ f\) in general.
- The domain of a composite function may be more restrictive than the domains of the individual functions.
- When simplifying expressions for combined functions, always consider the original domain restrictions.
Common Mistakes:
- Forgetting to exclude values that make a denominator zero in a quotient function.
- Assuming \(f \circ g = g \circ f\).
- Ignoring domain restrictions when simplifying rational expressions.
- Misapplying the order of operations when evaluating composite functions.
Problem-Solving Strategies:
- For arithmetic operations, perform the operation term-by-term and simplify.
- For composition, work from the inside out: evaluate the inner function first.
- Always determine the domain before simplifying, especially for quotients and composite functions with square roots.
- Use test points to verify domain restrictions.
Teacher-Led Examples:
Example 1: Arithmetic Operations
Given \(f(x) = 2x + 1\) and \(g(x) = x^2 - 4\), find \((f + g)(x)\), \((f - g)(x)\), \((f \cdot g)(x)\), and \(\left(\frac{f}{g}\right)(x)\). State the domain of each.
Solution:
- \((f + g)(x) = (2x + 1) + (x^2 - 4) = x^2 + 2x - 3\). Domain: all real numbers.
- \((f - g)(x) = (2x + 1) - (x^2 - 4) = -x^2 + 2x + 5\). Domain: all real numbers.
- \((f \cdot g)(x) = (2x + 1)(x^2 - 4) = 2x^3 + x^2 - 8x - 4\). Domain: all real numbers.
- \(\left(\frac{f}{g}\right)(x) = \frac{2x + 1}{x^2 - 4}\). Domain: all real numbers except \(x = 2\) and \(x = -2\), since \(g(x) = 0\) at those values.
Example 2: Composition of Functions
Given \(f(x) = 3x - 2\) and \(g(x) = x^2 + 1\), find \((f \circ g)(x)\) and \((g \circ f)(x)\). Evaluate \((f \circ g)(2)\).
Solution:
- \((f \circ g)(x) = f(g(x)) = f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1\).
- \((g \circ f)(x) = g(f(x)) = g(3x - 2) = (3x - 2)^2 + 1 = 9x^2 - 12x + 4 + 1 = 9x^2 - 12x + 5\).
- \((f \circ g)(2) = 3(2)^2 + 1 = 3(4) + 1 = 13\).
Note that \(f \circ g \neq g \circ f\).
Example 3: Domain of a Composite Function
Given \(f(x) = \sqrt{x}\) and \(g(x) = x - 3\), find \((f \circ g)(x)\) and its domain.
Solution:
- \((f \circ g)(x) = f(g(x)) = f(x - 3) = \sqrt{x - 3}\).
- The domain of \(g\) is all real numbers. The domain of \(f\) is \(x \geq 0\). For \(f(g(x))\) to be defined, we need \(g(x) \geq 0\), so \(x - 3 \geq 0\), which gives \(x \geq 3\). Thus, the domain of \(f \circ g\) is \([3, \infty)\).
Arithmetic Operations on Functions
Question: Given \(f(x) = 2x + 1\) and \(g(x) = x^2 - 4\), find \((f + g)(x)\), \((f - g)(x)\), \((f \cdot g)(x)\), and \(\left(\frac{f}{g}\right)(x)\). State the domain of each.
Solution: Step 1: Compute \((f + g)(x) = f(x) + g(x) = (2x + 1) + (x^2 - 4) = x^2 + 2x - 3\). The domain is all real numbers because both \(f\) and \(g\) are polynomials.
Step 2: Compute \((f - g)(x) = f(x) - g(x) = (2x + 1) - (x^2 - 4) = -x^2 + 2x + 5\). Domain: all real numbers.
Step 3: Compute \((f \cdot g)(x) = f(x) \cdot g(x) = (2x + 1)(x^2 - 4) = 2x^3 - 8x + x^2 - 4 = 2x^3 + x^2 - 8x - 4\). Domain: all real numbers.
Step 4: Compute \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{2x + 1}{x^2 - 4}\). The domain is all real numbers except where \(g(x) = 0\). Solve \(x^2 - 4 = 0 \Rightarrow x = \pm 2\). So domain: \(x \neq 2, -2\).
Composition of Functions
Question: Given \(f(x) = 3x - 2\) and \(g(x) = x^2 + 1\), find \((f \circ g)(x)\) and \((g \circ f)(x)\). Evaluate \((f \circ g)(2)\).
Solution: Step 1: Find \((f \circ g)(x) = f(g(x))\). Substitute \(g(x)\) into \(f\): \(f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1\).
Step 2: Find \((g \circ f)(x) = g(f(x))\). Substitute \(f(x)\) into \(g\): \(g(3x - 2) = (3x - 2)^2 + 1 = 9x^2 - 12x + 4 + 1 = 9x^2 - 12x + 5\).
Step 3: Evaluate \((f \circ g)(2)\). Using the formula: \(3(2)^2 + 1 = 3(4) + 1 = 13\). Alternatively, \(g(2) = 2^2 + 1 = 5\), then \(f(5) = 3(5) - 2 = 13\).
Note: \(f \circ g \neq g \circ f\).
Domain of a Composite Function
Question: Given \(f(x) = \sqrt{x}\) and \(g(x) = x - 3\), find \((f \circ g)(x)\) and its domain.
Solution: Step 1: Write the composition: \((f \circ g)(x) = f(g(x)) = f(x - 3) = \sqrt{x - 3}\).
Step 2: Determine the domain. The expression under the square root must be nonnegative: \(x - 3 \geq 0\), so \(x \geq 3\). Also, \(g(x)\) is defined for all real numbers, so no additional restrictions. Thus, the domain is \([3, \infty)\).
Step 3: Check with a test point: \(x = 3\) gives \(\sqrt{0} = 0\), valid. \(x = 2\) gives \(\sqrt{-1}\), not real, so excluded.
🤝 Guided Practice
Work with a partner to solve the following problems. Show all steps and be prepared to explain your reasoning. We will discuss as a class.
Given \(f(x) = x + 5\) and \(g(x) = 2x - 3\), find \((f - g)(x)\) and \((f \cdot g)(x)\). State the domain of each.
\((f - g)(x) = f(x) - g(x) = (x + 5) - (2x - 3) = x + 5 - 2x + 3 = -x + 8\). Domain: all real numbers.
\((f \cdot g)(x) = (x + 5)(2x - 3) = 2x^2 - 3x + 10x - 15 = 2x^2 + 7x - 15\). Domain: all real numbers.
Given \(f(x) = x^2\) and \(g(x) = x - 4\), find \((f \circ g)(x)\) and \((g \circ f)(x)\).
\((f \circ g)(x) = f(g(x)) = f(x - 4) = (x - 4)^2 = x^2 - 8x + 16\).
\((g \circ f)(x) = g(f(x)) = g(x^2) = x^2 - 4\).
Note that \(f \circ g \neq g \circ f\).
Given \(f(x) = \frac{1}{x}\) and \(g(x) = x + 2\), find \(\left(\frac{f}{g}\right)(x)\) and state its domain.
\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\frac{1}{x}}{x + 2} = \frac{1}{x(x + 2)}\).
Domain: \(x \neq 0\) and \(x \neq -2\), since the denominator cannot be zero. Also, \(x = 0\) is excluded from the domain of \(f\). So domain: all real numbers except \(0\) and \(-2\).
Given \(f(x) = \sqrt{x + 1}\) and \(g(x) = x^2\), find \((f \circ g)(x)\) and its domain. Explain your reasoning.
\((f \circ g)(x) = f(g(x)) = f(x^2) = \sqrt{x^2 + 1}\).
To find the domain, we need \(x^2 + 1 \geq 0\). Since \(x^2 \geq 0\) for all real \(x\), \(x^2 + 1 \geq 1 > 0\). Thus, the expression under the square root is always positive. Also, \(g(x) = x^2\) is defined for all real numbers. Therefore, the domain is all real numbers, \((-\infty, \infty)\).
Reasoning: The inner function \(g\) has domain all real numbers, and its output \(x^2\) is always nonnegative, so it always lies in the domain of \(f\) (which requires input \(\geq -1\)). Hence no additional restrictions.
🚀 Independent Practice
Solve the following problems independently. Show all work and clearly state domains where requested. These problems progress in difficulty and include exam-style questions.
Given \(f(x) = 3x + 2\) and \(g(x) = x^2 - 1\), find \((f + g)(x)\), \((f - g)(x)\), and \((f \cdot g)(x)\). State the domain of each.
\((f + g)(x) = (3x + 2) + (x^2 - 1) = x^2 + 3x + 1\). Domain: all real numbers.
\((f - g)(x) = (3x + 2) - (x^2 - 1) = -x^2 + 3x + 3\). Domain: all real numbers.
\((f \cdot g)(x) = (3x + 2)(x^2 - 1) = 3x^3 + 2x^2 - 3x - 2\). Domain: all real numbers.
Given \(f(x) = \frac{1}{x - 2}\) and \(g(x) = x + 3\), find \(\left(\frac{f}{g}\right)(x)\) and state its domain.
\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\frac{1}{x - 2}}{x + 3} = \frac{1}{(x - 2)(x + 3)}\).
Domain: \(x \neq 2\) (from \(f\)) and \(x \neq -3\) (from \(g(x) = 0\)). So domain: all real numbers except \(2\) and \(-3\).
Given \(f(x) = 2x - 1\) and \(g(x) = x^2 + 3x\), find \((f \circ g)(x)\) and \((g \circ f)(x)\). Simplify your answers.
\((f \circ g)(x) = f(g(x)) = f(x^2 + 3x) = 2(x^2 + 3x) - 1 = 2x^2 + 6x - 1\).
\((g \circ f)(x) = g(f(x)) = g(2x - 1) = (2x - 1)^2 + 3(2x - 1) = 4x^2 - 4x + 1 + 6x - 3 = 4x^2 + 2x - 2\).
Given \(f(x) = \sqrt{x - 2}\) and \(g(x) = 3x + 1\), find \((f \circ g)(x)\) and its domain.
\((f \circ g)(x) = f(g(x)) = f(3x + 1) = \sqrt{(3x + 1) - 2} = \sqrt{3x - 1}\).
Domain: need \(3x - 1 \geq 0 \Rightarrow x \geq \frac{1}{3}\). Also, \(g(x)\) is defined for all real numbers. So domain: \(\left[\frac{1}{3}, \infty\right)\).
Given \(f(x) = x^2 - 4\) and \(g(x) = \sqrt{x}\), find \((f \circ g)(x)\) and \((g \circ f)(x)\). State the domain of each composite function.
\((f \circ g)(x) = f(g(x)) = f(\sqrt{x}) = (\sqrt{x})^2 - 4 = x - 4\).
Domain: \(x \geq 0\) because \(g(x) = \sqrt{x}\) requires \(x \geq 0\). Also, \(g(x)\) output is always \(\geq 0\), which is in the domain of \(f\) (all real numbers). So domain: \([0, \infty)\).
\((g \circ f)(x) = g(f(x)) = g(x^2 - 4) = \sqrt{x^2 - 4}\).
Domain: need \(x^2 - 4 \geq 0 \Rightarrow x^2 \geq 4 \Rightarrow |x| \geq 2\). So domain: \((-\infty, -2] \cup [2, \infty)\).
The population of a town is modeled by \(P(t) = 5000 + 200t\), where \(t\) is the number of years since 2020. The number of schools needed is modeled by \(S(P) = \frac{P}{1000} + 2\), where \(P\) is the population. Find \((S \circ P)(t)\) and interpret its meaning. How many schools are needed in 2030?
\((S \circ P)(t) = S(P(t)) = S(5000 + 200t) = \frac{5000 + 200t}{1000} + 2 = 5 + 0.2t + 2 = 7 + 0.2t\).
This function gives the number of schools needed \(t\) years after 2020.
In 2030, \(t = 10\). \((S \circ P)(10) = 7 + 0.2(10) = 7 + 2 = 9\). So 9 schools are needed in 2030.
Given \(f(x) = \frac{2x + 1}{x - 3}\) and \(g(x) = x + 2\), find \((f \circ g)(x)\) and state its domain.
\((f \circ g)(x) = f(g(x)) = f(x + 2) = \frac{2(x + 2) + 1}{(x + 2) - 3} = \frac{2x + 4 + 1}{x - 1} = \frac{2x + 5}{x - 1}\).
Domain: \(x \neq 1\) (from denominator of composite). Also, \(g(x) = x + 2\) is defined for all real numbers, but we must ensure \(g(x)\) is in the domain of \(f\). The domain of \(f\) is \(x \neq 3\). So we need \(g(x) \neq 3 \Rightarrow x + 2 \neq 3 \Rightarrow x \neq 1\). This matches the denominator restriction. Thus domain: all real numbers except \(1\).
Let \(f(x) = \sqrt{x}\) and \(g(x) = x^2 - 9\). Find \((f \circ g)(x)\) and its domain. Then find \((g \circ f)(x)\) and its domain. Are the two composite functions equal? Explain.
\((f \circ g)(x) = f(g(x)) = f(x^2 - 9) = \sqrt{x^2 - 9}\).
Domain: need \(x^2 - 9 \geq 0 \Rightarrow x^2 \geq 9 \Rightarrow |x| \geq 3\). So domain: \((-\infty, -3] \cup [3, \infty)\).
\((g \circ f)(x) = g(f(x)) = g(\sqrt{x}) = (\sqrt{x})^2 - 9 = x - 9\).
Domain: \(x \geq 0\) because \(f(x) = \sqrt{x}\) requires \(x \geq 0\). Also, \(f(x)\) output is \(\geq 0\), which is in the domain of \(g\) (all real numbers). So domain: \([0, \infty)\).
The two composite functions are not equal because they have different domains and different formulas. \(f \circ g\) has domain \((-\infty, -3] \cup [3, \infty)\), while \(g \circ f\) has domain \([0, \infty)\).
🎫 Exit Ticket
Given \(f(x) = 2x + 3\) and \(g(x) = x^2 - 1\), find \((f \circ g)(x)\) and evaluate \((f \circ g)(2)\).
\((f \circ g)(x) = f(g(x)) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1\).
\((f \circ g)(2) = 2(2)^2 + 1 = 2(4) + 1 = 9\).
Given \(f(x) = \frac{1}{x}\) and \(g(x) = x - 4\), find \(\left(\frac{f}{g}\right)(x)\) and state its domain.
\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\frac{1}{x}}{x - 4} = \frac{1}{x(x - 4)}\).
Domain: \(x \neq 0\) (from \(f\)) and \(x \neq 4\) (from \(g(x) = 0\)). So domain: all real numbers except \(0\) and \(4\).
🧑🏫 Teacher Reflection & Differentiation
Common Misconceptions
- Students may confuse function composition with multiplication, writing \((f \circ g)(x)\) as \(f(x) \cdot g(x)\).
- Students may assume that \(f \circ g = g \circ f\), not realizing that composition is not commutative.
- Students may forget to exclude values that make a denominator zero when finding the domain of a quotient function.
- Students may ignore domain restrictions when simplifying composite functions, especially when square roots are involved.
- Students may incorrectly evaluate composite functions by substituting into the wrong function first.
Supporting Struggling Students
Provide students with a step-by-step graphic organizer for function operations and composition. Use color-coding to distinguish between \(f\) and \(g\). Offer additional practice with simpler linear functions before moving to rational and radical functions. Use think-alouds to model the process of finding domains.
Challenging Advanced Students
Challenge students with problems involving three functions, such as finding \((f \circ g \circ h)(x)\). Ask students to explore conditions under which \(f \circ g = g \circ f\). Have students create real-world problems that require composition of functions and present their solutions to the class.
Suggested Re-teaching Strategy
If students struggle on the Exit Ticket, reteach the concept of composition by using a function machine analogy: the output of one machine becomes the input of another. Use a table of values to demonstrate step-by-step evaluation. Re-emphasize the difference between \(f \circ g\) and \(f \cdot g\). Provide targeted practice on finding domains of composite functions, especially those involving square roots and rational expressions. Use peer tutoring or small group instruction to address specific misconceptions.