AP Precalculus: Function Concepts & Formulas
Master the essential function concepts, formulas, and techniques for AP Precalculus success
📚 About This Reference Sheet
This comprehensive guide covers the fundamental function concepts tested on the AP Precalculus exam. Each section includes clear definitions, essential formulas, key points to remember, and worked examples to help you master these critical skills. Use this as your go-to reference for studying and exam preparation.
1 Domain and Range
Domain: The set of all valid input values (\(x\)) for which the function \(f(x)\) is defined.
Range: The set of all possible output values \(f(x)\) that the function can produce.
Key Rules for Finding Domain
- Rational functions: Exclude values that make the denominator equal to zero
- Square root functions: The radicand (expression under the radical) must be \(\geq 0\)
- Logarithmic functions: The argument must be \(> 0\)
- Polynomial functions: Domain is always all real numbers \((-\infty, \infty)\)
Ways to Express Domain and Range
- Interval notation: \([-1, \infty)\), \((-\infty, 3) \cup (3, \infty)\)
- Set-builder notation: \(\{x \mid x \geq -1\}\)
- Inequality notation: \(x \geq -1\) or \(x \neq 0\)
\(f(x) = x^2 - 1\)
Domain: \((-\infty, \infty)\) — all real numbers
Range: \([-1, \infty)\) — minimum value is -1
\(f(x) = \frac{1}{x}\)
Domain: \(x \neq 0\) or \((-\infty, 0) \cup (0, \infty)\)
\(f(x) = \sqrt{x-5}\)
Domain: \(x \geq 5\) or \([5, \infty)\)
To find the range of a function, consider the possible outputs. For quadratic functions like \(f(x) = ax^2 + bx + c\), the vertex gives you the minimum (if \(a > 0\)) or maximum (if \(a < 0\)) value of the range.
2 Identifying Functions
A function is a relation that assigns exactly one output to each input. Every value of \(x\) in the domain must have one and only one corresponding \(y\) value.
How to Identify Functions
- From a graph: Apply the vertical line test — draw vertical lines across the graph
- From a table: Check if each \(x\)-value has only one \(y\)-value
- From an equation: Solve for \(y\) — if you get more than one \(y\) for any \(x\), it's not a function
- From a set of ordered pairs: No two pairs can have the same first element with different second elements
✓ \(f(x) = x^2\) — IS a function (passes vertical line test)
✓ \(f(x) = 2x + 3\) — IS a function (each \(x\) gives one \(y\))
✗ \(x^2 + y^2 = 1\) (circle) — NOT a function (fails vertical line test)
✗ \(y = \pm\sqrt{x}\) — NOT a function (two \(y\) values for each positive \(x\))
3 Evaluating Functions
To evaluate a function means to find the output value when a specific input value is substituted into the function expression.
Read as "f of a"
Steps to Evaluate Functions
- Identify the input value (what goes inside the parentheses)
- Substitute the input value for every instance of \(x\) in the function
- Simplify the expression using order of operations
- For expressions like \(f(a+b)\), replace \(x\) with the entire expression \((a+b)\)
Given: \(f(x) = 2x + 3\)
\(f(4) = 2(4) + 3 = 8 + 3 = 11\)
\(f(-2) = 2(-2) + 3 = -4 + 3 = -1\)
\(f(a+b) = 2(a+b) + 3 = 2a + 2b + 3\)
Given: \(g(x) = x^2 - 4x + 1\)
\(g(3) = (3)^2 - 4(3) + 1 = 9 - 12 + 1 = -2\)
\(g(0) = (0)^2 - 4(0) + 1 = 1\)
When evaluating \(f(x+h)\), don't forget to substitute for every \(x\) in the expression. For \(f(x) = x^2\), we get \(f(x+h) = (x+h)^2 = x^2 + 2xh + h^2\), not \(x^2 + h\).
4 Finding Values from Function Graphs
Reading function values from a graph involves locating the input value on the \(x\)-axis and finding the corresponding output value on the \(y\)-axis.
How to Read Values from a Graph
- To find \(f(a)\): Locate \(x = a\) on the horizontal axis, move vertically to the graph, then read the \(y\)-coordinate
- To solve \(f(x) = b\): Locate \(y = b\) on the vertical axis, move horizontally to the graph, then read all \(x\)-coordinates where they intersect
- To find domain: Identify all \(x\)-values where the graph exists
- To find range: Identify all \(y\)-values that the graph reaches
If a graph passes through the point \((2, 5)\), then \(f(2) = 5\).
If the graph intersects \(y = 3\) at \(x = -1\) and \(x = 4\), then the solutions to \(f(x) = 3\) are \(x = -1\) and \(x = 4\).
Key points to identify on any function graph: \(x\)-intercepts (where \(f(x) = 0\)), \(y\)-intercept (the value of \(f(0)\)), maximum/minimum points, and any asymptotes or holes.
5 Completing Tables for Functions
A function table organizes input-output pairs. To complete a table, substitute each \(x\)-value into the function and calculate the corresponding \(f(x)\) value.
Steps to Complete a Function Table
- Identify the function rule or equation
- For each given \(x\)-value, substitute into the function
- Calculate and record the output \(f(x)\)
- If given \(f(x)\), solve for the corresponding \(x\)
| \(x\) | Calculation | \(f(x)\) |
|---|---|---|
| -1 | \(2(-1) + 1 = -1\) | -1 |
| 0 | \(2(0) + 1 = 1\) | 1 |
| 1 | \(2(1) + 1 = 3\) | 3 |
| 2 | \(2(2) + 1 = 5\) | 5 |
| 3 | \(2(3) + 1 = 7\) | 7 |
6 Interpreting Graphs in Context
In real-world applications, graphs represent relationships between quantities. Interpreting these graphs means connecting graph features (intercepts, slopes, maxima, minima) to their real-world meaning.
Key Graph Features and Their Meanings
- Y-intercept: Initial value, starting point, or value when \(x = 0\)
- X-intercept: When the quantity equals zero (break-even point, ground level, etc.)
- Maximum point: Peak value, highest point, greatest profit, etc.
- Minimum point: Lowest value, deepest point, minimum cost, etc.
- Increasing intervals: Where the quantity is growing or rising
- Decreasing intervals: Where the quantity is shrinking or falling
- Slope (rate of change): How fast the quantity changes per unit of input
Context: A graph shows a company's profit (in thousands of dollars) over months since opening.
• Y-intercept at -50: The company started with a $50,000 debt/loss
• X-intercept at month 6: The company broke even after 6 months
• Maximum at (12, 80): Peak profit of $80,000 occurred at month 12
• Increasing from 0 to 12: Profit was growing during the first year
7 Function Operations (+, −, ×, ÷)
Functions can be combined using arithmetic operations to create new functions. The resulting function's domain is the intersection of the original domains (with additional restrictions for division).
The Four Function Operations
For \(\left(\frac{f}{g}\right)(x)\), the domain excludes all values where \(g(x) = 0\).
Given: \(f(x) = x^2\) and \(g(x) = x + 1\)
\((f + g)(x) = x^2 + (x + 1) = x^2 + x + 1\)
\((f - g)(x) = x^2 - (x + 1) = x^2 - x - 1\)
\((f \cdot g)(x) = x^2 \cdot (x + 1) = x^3 + x^2\)
\(\left(\frac{f}{g}\right)(x) = \frac{x^2}{x + 1}\), where \(x \neq -1\)
8 Composition of Functions
Function composition creates a new function by using the output of one function as the input of another. The notation \((f \circ g)(x)\) means "apply \(g\) first, then apply \(f\) to the result."
Read as "f composed with g of x" or "f of g of x"
Steps for Function Composition
- Start with the "inner" function (the one closest to \(x\))
- Evaluate or substitute the inner function's expression
- Use that result as the input for the "outer" function
- Simplify the final expression
\((f \circ g)(x) \neq (g \circ f)(x)\) in general. Function composition is not commutative.
Given: \(f(x) = 2x + 1\) and \(g(x) = x^2\)
Find \((f \circ g)(x) = f(g(x))\):
\(f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1\)
Find \((g \circ f)(x) = g(f(x))\):
\(g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1\)
Notice: \(2x^2 + 1 \neq 4x^2 + 4x + 1\), confirming that order matters!
The domain of \(f \circ g\) includes all \(x\) values in the domain of \(g\) such that \(g(x)\) is in the domain of \(f\). You must check both conditions!