Learning Objectives
Evaluate a function and distinguish f(a) from an equation to solve.
Determine algebraic and contextual domain restrictions.
Find the range of a simple function on a stated domain.
Prerequisites: algebraic substitution, inequalities, and square roots. This section develops shared function foundations without claiming identical AA/AI or SL/HL coverage.
1. A Function Assigns One Output to Each Allowed Input
A function maps each input in its domain to exactly one output. Different inputs may share an output. For example, f(x) = x² sends both 2 and −2 to 4, which is permitted. A rule assigning two outputs to the same allowed input would not define a function.
Worked Example: Evaluation and Solving
Let f(x) = 2x² − 3. To find f(4), substitute 4 for x: f(4) = 2(16) − 3 = 29.
To solve f(x) = 5, set 2x² − 3 = 5. Then x² = 4, so x = −2 or x = 2.
Evaluation finds an output from a specified input. Solving f(x) = 5 finds the inputs producing a specified output.
2. The Domain Lists Allowed Inputs
For a real-valued formula, denominators must be nonzero and expressions under even roots must be nonnegative. A real logarithm requires a positive argument. Combine restrictions by intersection. A separately stated domain can narrow the set further.
Worked Example: Two Restrictions
Find the domain of g(x) = √(x + 2)/(x − 1). The square root requires x ≥ −2, and the denominator requires x ≠ 1.
Therefore the domain is [−2, 1) ∪ (1, ∞). The endpoint −2 is included because √0 is defined and the denominator there is nonzero.
Do not include x = 1 merely because the square root works there; every part of the expression must be defined.
3. The Range Lists Outputs Actually Produced
A formula's range depends on its domain. For f(x) = x² on all real x, the range is [0, ∞). If the domain is restricted to 2 ≤ x ≤ 5, the range becomes [4, 25]. The same formula can therefore describe different functions when its allowed inputs change.
Worked Example: A Turning Point inside the Domain
Find the range of h(x) = (x − 2)² + 1 on 0 ≤ x ≤ 5. The squared term is smallest at x = 2, which lies in the domain, so the minimum is 1.
Check both endpoints: h(0) = 5 and h(5) = 10. The maximum is 10.
The function takes every value between these extremes on the interval, so its range is [1, 10]. Checking endpoints alone would have missed the minimum.
4. Context Can Restrict a Formula
If C(n) = 4n + 7 is a cost for n items, the algebraic expression is defined for all real n. But item counts may restrict n to nonnegative integers, or to 0 through a stock limit. State whether a model treats quantities as discrete counts or continuous measurements.
Common Mistakes
The notation f(x) means a function value, not f multiplied by x.
A valid algebraic input may still be impossible in the context.
When finding a range on an interval, check relevant turning points as well as endpoints.
5. Practice
1. If f(x) = 3x − 4, find f(−2) and solve f(x) = 8.
2. Find the domain of 1/√(5 − x).
3. Find the range of x² on −2 ≤ x ≤ 3.
Worked Solutions
1. f(−2) = −6 − 4 = −10. Solving 3x − 4 = 8 gives x = 4.
2. The root is in the denominator, so 5 − x must be strictly positive. The domain is x < 5, or (−∞, 5).
3. The minimum is 0 at x = 0. The endpoint values are 4 and 9, so the maximum is 9. The range is [0, 9].
6. Summary
- A function pairs each allowed input with one output.
- Evaluation and solving reverse which quantity is given.
- Find the domain before interpreting the range.
- Use the stated context and interval when reporting allowable values.