Difficulty: Introductory | Prerequisites: Basic algebra (factoring, simplifying fractions, square roots).
This material sits at the very start of a business calculus course. It covers how to evaluate functions (including piecewise ones), how to find where a function is defined, and the groundwork ideas of limits and continuity that the rest of the course builds on. If you are comfortable substituting values into expressions and simplifying with exponents and roots, you have what you need. If those feel shaky, revisit your algebra foundations first.
A function is a rule that takes an input and gives back exactly one output. To evaluate one, plug in the number and simplify. Piecewise functions use different rules depending on the input value, so check which piece applies before you substitute. The domain of a function is every input value for which the output is a real number, meaning you exclude values that cause division by zero or a negative under a square root.
Function
A rule that assigns exactly one output to each input. Written as f(x), where x is the input and f(x) is the output.
In simple terms, this means: you feed in a number, the rule does something to it, and you get one answer back.
Function evaluation
The process of substituting a specific value for the variable in a function's formula and simplifying.
Think of it as: replacing the letter with the number, then doing the arithmetic.
Piecewise function
A function defined by two or more expressions, each applying on a different part of the domain.
In simple terms, this means: the formula changes depending on the value of x. You check which condition your input satisfies, then use the corresponding rule.
Domain
The set of all input values for which a function produces a real-number output.
Think of it as: every x-value you are allowed to plug in without breaking the maths (no dividing by zero, no square roots of negatives).
Interval notation
A shorthand for writing sets of numbers. Round brackets ( ) mean the endpoint is excluded; square brackets [ ] mean it is included. For example, [5, ∞) means "all numbers from 5 onwards, including 5."
Square root function
A function of the form g(x) = √(expression). The expression under the radical must be zero or positive for the output to be a real number.
In simple terms, this means: you cannot take the square root of a negative number in this course, so the bit inside the √ has to be ≥ 0.
Direct substitution with polynomial functions
Given f(x) = x² + 5, find f(3):
Replace x with 3: f(3) = 3² + 5 = 9 + 5 = 14.
The key step is squaring first, then adding. Follow order of operations.
Evaluating radical (root) expressions
Given g(t) = (t – 2)^(1/2), the exponent 1/2 means a square root.
g(27) = (27 – 2)^(1/2) = 25^(1/2) = √25 = 5.
g(5) = (5 – 2)^(1/2) = 3^(1/2) = √3 (leave as a surd, do not approximate).
g(2) = (2 – 2)^(1/2) = 0^(1/2) = 0. Zero under a square root is fine.
g(1) = (1 – 2)^(1/2) = (–1)^(1/2) = √(–1), which is not a real number. The function is undefined at t = 1.
Takeaway: before simplifying, check whether the expression under the radical is negative. If it is, stop and state "not real."
A piecewise function has different formulas for different ranges of x. The definition spells out each rule and the condition that goes with it.
Example from the activity:
f(x) = 1/(x – 1) if x < 1 (and x ≠ 1), or f(x) = 3x² + 2 if x ≥ 1.
How to evaluate:
Determine which piece applies by checking the condition on x.
Substitute into the correct formula only.
Worked examples:
f(–1/2): since –1/2 < 1, use the first piece.
f(–1/2) = 1/((–1/2) – 1) = 1/(–3/2) = –2/3.
f(1): since 1 ≥ 1, use the second piece.
f(1) = 3(1)² + 2 = 3 + 2 = 5.
f(2): since 2 ≥ 1, use the second piece.
f(2) = 3(2)² + 2 = 3(4) + 2 = 12 + 2 = 14.
Common error: students plug into the wrong piece because they check the condition carelessly. Always identify the correct branch before doing any arithmetic.
Two rules govern most domain questions at this level:
You cannot divide by zero. If the function has a denominator, set it ≠ 0 and exclude that value.
You cannot take the square root of a negative number. If the function has a radical, set the expression inside ≥ 0.
When both a radical and a denominator are present, apply both rules.
Example 1: f(x) = 1/(x – 3)
Set x – 3 ≠ 0, so x ≠ 3.
Domain: (–∞, 3) ∪ (3, ∞).
Example 2: g(x) = √(x – 5)
Set x – 5 ≥ 0, so x ≥ 5.
Domain: [5, ∞).
Example 3: f(x) = √(x + 8) / (x – 2)
Radical condition: x + 8 ≥ 0, so x ≥ –8.
Denominator condition: x – 2 ≠ 0, so x ≠ 2.
Combine: x ≥ –8 and x ≠ 2.
Domain: [–8, 2) ∪ (2, ∞).
Writing in interval notation:
Use [ ] when the endpoint is included (≤ or ≥).
Use ( ) when it is excluded (< or >, or a hole from division by zero).
∞ and –∞ always get a round bracket, because infinity is not a number you reach.
f(x) = x^2 + 5g(t) = (t - 2)^{\frac{1}{2}} = \sqrt{t - 2}Piecewise function:
f(x) = \begin{cases} \dfrac{1}{x-1} & \text{if } x < 1,\, x \neq 1 \\ 3x^2 + 2 & \text{if } x \geq 1 \end{cases}Domain restrictions to remember:
Denominator ≠ 0.
Radicand (expression under √) ≥ 0.
Piecewise functions show up constantly in business contexts. Tax brackets are piecewise: the rate changes depending on income. Shipping costs often follow a piecewise model where the price per kilogram drops above a certain weight threshold.
Domain restrictions matter whenever a model involves division (cost per unit breaks down at zero units produced) or square roots (standard deviation formulas, for instance, require non-negative inputs).
Students often think x^(1/2) means "divide x by 2." It does not. The exponent 1/2 means the square root of x.
Students forget that √0 = 0 and is perfectly valid. Zero under a radical is not the same as a negative under a radical.
When finding domains, students sometimes exclude x = 0 by reflex. Only exclude values that actually cause division by zero or a negative radicand, not zero itself.
With piecewise functions, students sometimes evaluate both pieces and then pick the larger answer. You must use only the one piece whose condition the input satisfies.
⚠️ Function evaluation with substitution is tested on nearly every exam. Expect to evaluate polynomial, radical, and piecewise functions under time pressure.
⚠️ Domain questions in interval notation are a staple of business calculus exams. You will be expected to write domains using ( ), [ ], and ∪ correctly.
⚠️ Piecewise function evaluation is a favourite for tricky exam questions. The examiner is testing whether you read the conditions carefully.
⚠️ When a function involves both a radical and a fraction, expect a domain question that requires you to combine both restrictions. This is where marks are most commonly dropped.
True or False: (–4)^(1/2) is a real number.
Fill in the blank: The domain of f(x) = 1/(x + 7) is all real numbers except x = ___.
True or False: For the piecewise function f(x) = {x² if x < 0, 2x + 1 if x ≥ 0}, f(0) = 0.
Fill in the blank: In interval notation, x > 3 is written as (3, ___).
True or False: √0 is undefined.
Answers: 1. False. 2. –7. 3. False (f(0) = 2(0) + 1 = 1, using the second piece). 4. ∞). 5. False (√0 = 0).
Q: If f(x) = 2x³ – x + 4, what is f(–1)?
A: f(–1) = 2(–1)³ – (–1) + 4 = 2(–1) + 1 + 4 = –2 + 1 + 4 = 3.
Q: Find the domain of h(x) = √(3 – x).
A: Set 3 – x ≥ 0, so x ≤ 3. Domain: (–∞, 3].
Q: Given f(x) = {5x + 1 if x < 2, x² – 3 if x ≥ 2}, find f(2) and f(0).
A: f(2): since 2 ≥ 2, use x² – 3. f(2) = 4 – 3 = 1. f(0): since 0 < 2, use 5x + 1. f(0) = 0 + 1 = 1.
Q: What is the domain of g(x) = √(x + 1) / (x – 4)?
A: Radical: x + 1 ≥ 0, so x ≥ –1. Denominator: x ≠ 4. Domain: [–1, 4) ∪ (4, ∞).
Q: Explain why g(t) = (t – 2)^(1/2) is undefined at t = 1.
A: At t = 1, the expression under the radical is 1 – 2 = –1. The square root of a negative number is not a real number, so g(1) does not exist in the reals.
Function evaluation is the foundation for derivatives. When you learn the difference quotient (the formal definition of a derivative), you will evaluate f(x + h) and f(x) repeatedly, so fluency here saves time later.
Domain analysis connects directly to continuity. A function cannot be continuous at a point outside its domain. When the course moves to limits and continuity, you will revisit domains to identify where limits may fail.
Piecewise functions reappear in business applications such as cost functions (fixed cost below a threshold, variable cost above it) and tax models.
Function evaluation, polynomial functions, piecewise functions, piecewise-defined functions, domain of a function, interval notation, radical functions, square root functions, exponent 1/2, fractional exponents, domain restrictions, division by zero, radicand, business calculus, Calculus for Business, FIU, Florida International University, limits and continuity prerequisites, direct substitution, function notation, f(x), evaluating functions at a point.