College Algebra

Functions and Notation (Rigorous Treatment)

Last updated: March 2026 · Advanced
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In Algebra 2, you worked with functions as rules that transform inputs into outputs. College Algebra demands more precision. Here we define functions formally using set theory, master the difference quotient that leads to calculus, and evaluate functions defined by tables, graphs, formulas, and piecewise rules.

Formal Definition of a Function

A function from set AA to set BB is a set of ordered pairs (x,y)(x, y) where:

  1. Every element xx in AA appears as the first coordinate of exactly one pair.
  2. No two distinct pairs share the same first coordinate.

In set notation, a function f:Aβ†’Bf: A \to B satisfies: for every x∈Ax \in A, there exists a unique y∈By \in B such that (x,y)∈f(x, y) \in f.

f={(x,y)∣x∈A,β€…β€Šy∈B,β€…β€ŠandΒ eachΒ xΒ mapsΒ toΒ exactlyΒ oneΒ y}f = \{(x, y) \mid x \in A,\; y \in B,\; \text{and each } x \text{ maps to exactly one } y\}

Example: The set {(1,3),(2,5),(3,7),(4,9)}\{(1, 3), (2, 5), (3, 7), (4, 9)\} is a function because each first coordinate appears once. The set {(1,3),(1,5),(2,7)}\{(1, 3), (1, 5), (2, 7)\} is not a function because the input 11 maps to two different outputs.

Domain and Range

  • The domain is the set of all first coordinates: Dom(f)={x∣(x,y)∈f}\text{Dom}(f) = \{x \mid (x, y) \in f\}
  • The range is the set of all second coordinates that are actually produced: Range(f)={y∣(x,y)∈f}\text{Range}(f) = \{y \mid (x, y) \in f\}

The domain is the set of all valid inputs; the range is the set of all actual outputs. A common mistake is to confuse the codomain (the target set BB) with the range. The codomain is where outputs could land; the range is where they actually land.

Function Notation Review

When we write f(x)=3x+2f(x) = 3x + 2, we are naming the function ff and declaring the rule: take any input xx, multiply by 3, and add 2. The expression f(x)f(x) is read β€œf of x” and represents the output when xx is the input.

Key evaluation patterns:

  • f(5)=3(5)+2=17f(5) = 3(5) + 2 = 17
  • f(βˆ’2)=3(βˆ’2)+2=βˆ’4f(-2) = 3(-2) + 2 = -4
  • f(a)=3a+2f(a) = 3a + 2
  • f(a+1)=3(a+1)+2=3a+3+2=3a+5f(a + 1) = 3(a + 1) + 2 = 3a + 3 + 2 = 3a + 5
  • f(x2)=3x2+2f(x^2) = 3x^2 + 2

The parentheses in f(x)f(x) do not mean multiplication. This is a notation convention β€” f(x)f(x) names the output of the function ff at input xx.

Evaluating More Complex Expressions

Given g(x)=x2βˆ’4x+1g(x) = x^2 - 4x + 1, find g(3)g(3), g(βˆ’1)g(-1), and g(2t)g(2t).

g(3)=(3)2βˆ’4(3)+1=9βˆ’12+1=βˆ’2g(3) = (3)^2 - 4(3) + 1 = 9 - 12 + 1 = -2

g(βˆ’1)=(βˆ’1)2βˆ’4(βˆ’1)+1=1+4+1=6g(-1) = (-1)^2 - 4(-1) + 1 = 1 + 4 + 1 = 6

g(2t)=(2t)2βˆ’4(2t)+1=4t2βˆ’8t+1g(2t) = (2t)^2 - 4(2t) + 1 = 4t^2 - 8t + 1

Notice how the entire input expression replaces every instance of xx in the formula, and parentheses protect the substitution from sign and order errors.

The Difference Quotient

The difference quotient is the expression:

f(x+h)βˆ’f(x)h,hβ‰ 0\frac{f(x + h) - f(x)}{h}, \quad h \neq 0

This measures the average rate of change of ff between xx and x+hx + h. It is the slope of the secant line through the points (x,f(x))(x, f(x)) and (x+h,f(x+h))(x + h, f(x + h)). In calculus, taking the limit as h→0h \to 0 produces the derivative — the instantaneous rate of change.

Worked Example 1: Difference Quotient for a Quadratic

Find and simplify the difference quotient for f(x)=x2+3xf(x) = x^2 + 3x.

Step 1: Compute f(x+h)f(x + h).

f(x+h)=(x+h)2+3(x+h)=x2+2xh+h2+3x+3hf(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h

Step 2: Compute f(x+h)βˆ’f(x)f(x + h) - f(x).

f(x+h)βˆ’f(x)=(x2+2xh+h2+3x+3h)βˆ’(x2+3x)=2xh+h2+3hf(x + h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h) - (x^2 + 3x) = 2xh + h^2 + 3h

Step 3: Divide by hh.

f(x+h)βˆ’f(x)h=2xh+h2+3hh=h(2x+h+3)h=2x+h+3\frac{f(x + h) - f(x)}{h} = \frac{2xh + h^2 + 3h}{h} = \frac{h(2x + h + 3)}{h} = 2x + h + 3

The simplified difference quotient is 2x+h+32x + h + 3.

Worked Example 2: Difference Quotient for a Rational Function

Find and simplify the difference quotient for f(x)=1xf(x) = \frac{1}{x}.

Step 1: Compute f(x+h)βˆ’f(x)f(x + h) - f(x).

1x+hβˆ’1x=xβˆ’(x+h)x(x+h)=βˆ’hx(x+h)\frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)}

Step 2: Divide by hh.

f(x+h)βˆ’f(x)h=βˆ’hx(x+h)β‹…1h=βˆ’1x(x+h)\frac{f(x + h) - f(x)}{h} = \frac{-h}{x(x + h)} \cdot \frac{1}{h} = \frac{-1}{x(x + h)}

The simplified difference quotient is βˆ’1x(x+h)\frac{-1}{x(x + h)}.

Functions Defined Multiple Ways

By Formula

Most functions in algebra are defined by formulas like f(x)=2x2βˆ’5f(x) = 2x^2 - 5. The domain is all real numbers unless the formula involves division (denominator cannot be zero) or even roots (radicand must be non-negative).

By Table

xxf(x)f(x)
βˆ’2-277
0033
1111
44βˆ’5-5

From this table: f(0)=3f(0) = 3, f(4)=βˆ’5f(4) = -5, and the domain is {βˆ’2,0,1,4}\{-2, 0, 1, 4\}.

By Graph (Vertical Line Test)

A graph in the coordinate plane represents a function if and only if every vertical line intersects the graph at most once. This is the vertical line test. If any vertical line crosses the graph twice, there exists an input with two outputs, violating the function definition.

By Verbal Description

Engineering example: The stress Οƒ\sigma in a beam depends on the applied force FF and the cross-sectional area AA: Οƒ(F)=FA\sigma(F) = \frac{F}{A}, where AA is a constant for a given beam. This verbal relationship defines a function.

Piecewise Functions: First Look

A piecewise-defined function uses different formulas on different parts of the domain.

f(x)={x2ifΒ x≀13xβˆ’2ifΒ x>1f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 3x - 2 & \text{if } x > 1 \end{cases}

To evaluate:

  • f(βˆ’3)=(βˆ’3)2=9f(-3) = (-3)^2 = 9 (since βˆ’3≀1-3 \leq 1, use the first piece)
  • f(1)=(1)2=1f(1) = (1)^2 = 1 (since 1≀11 \leq 1, use the first piece)
  • f(4)=3(4)βˆ’2=10f(4) = 3(4) - 2 = 10 (since 4>14 > 1, use the second piece)

Retail application: A store offers a bulk discount. The cost function might be:

C(n)={5nifΒ 1≀n≀1004nifΒ n>100C(n) = \begin{cases} 5n & \text{if } 1 \leq n \leq 100 \\ 4n & \text{if } n > 100 \end{cases}

where nn is the number of items. At 80 items the cost is C(80)=5(80)=400C(80) = 5(80) = 400. At 150 items the cost is C(150)=4(150)=600C(150) = 4(150) = 600.

Real-World Application: Engineering Cost Analysis

An engineering firm models the total cost of a project as C(x)=0.02x2+500x+12000C(x) = 0.02x^2 + 500x + 12000, where xx is the number of labor hours. The difference quotient tells the firm how fast costs are changing:

C(x+h)βˆ’C(x)h=0.02(x+h)2+500(x+h)+12000βˆ’0.02x2βˆ’500xβˆ’12000h\frac{C(x + h) - C(x)}{h} = \frac{0.02(x+h)^2 + 500(x+h) + 12000 - 0.02x^2 - 500x - 12000}{h}

Expanding and simplifying:

=0.04xh+0.02h2+500hh=0.04x+0.02h+500= \frac{0.04xh + 0.02h^2 + 500h}{h} = 0.04x + 0.02h + 500

At x=100x = 100 hours with h=1h = 1, the approximate marginal cost is 0.04(100)+0.02(1)+500=504.020.04(100) + 0.02(1) + 500 = 504.02 per additional hour.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Determine whether {(2,4),(3,6),(2,8),(5,10)}\{(2, 4), (3, 6), (2, 8), (5, 10)\} is a function. Explain why or why not.

This is not a function because the input 22 maps to two different outputs (44 and 88). A function requires each input to produce exactly one output.

Answer: Not a function β€” the input 22 has two outputs.

Problem 2: Given f(x)=2x2βˆ’x+3f(x) = 2x^2 - x + 3, find f(βˆ’2)f(-2) and f(a+1)f(a + 1).

f(βˆ’2)=2(βˆ’2)2βˆ’(βˆ’2)+3=2(4)+2+3=13f(-2) = 2(-2)^2 - (-2) + 3 = 2(4) + 2 + 3 = 13

f(a+1)=2(a+1)2βˆ’(a+1)+3=2(a2+2a+1)βˆ’aβˆ’1+3=2a2+4a+2βˆ’a+2=2a2+3a+4f(a + 1) = 2(a + 1)^2 - (a + 1) + 3 = 2(a^2 + 2a + 1) - a - 1 + 3 = 2a^2 + 4a + 2 - a + 2 = 2a^2 + 3a + 4

Answer: f(βˆ’2)=13f(-2) = 13 and f(a+1)=2a2+3a+4f(a + 1) = 2a^2 + 3a + 4.

Problem 3: Find and simplify the difference quotient for f(x)=3x2βˆ’2xf(x) = 3x^2 - 2x.

f(x+h)=3(x+h)2βˆ’2(x+h)=3x2+6xh+3h2βˆ’2xβˆ’2hf(x + h) = 3(x + h)^2 - 2(x + h) = 3x^2 + 6xh + 3h^2 - 2x - 2h

f(x+h)βˆ’f(x)=6xh+3h2βˆ’2hf(x + h) - f(x) = 6xh + 3h^2 - 2h

f(x+h)βˆ’f(x)h=h(6x+3hβˆ’2)h=6x+3hβˆ’2\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h - 2)}{h} = 6x + 3h - 2

Answer: The simplified difference quotient is 6x+3hβˆ’26x + 3h - 2.

Problem 4: Find and simplify the difference quotient for f(x)=2x+1f(x) = \frac{2}{x + 1}.

f(x+h)βˆ’f(x)=2x+h+1βˆ’2x+1=2(x+1)βˆ’2(x+h+1)(x+h+1)(x+1)=βˆ’2h(x+h+1)(x+1)f(x + h) - f(x) = \frac{2}{x + h + 1} - \frac{2}{x + 1} = \frac{2(x + 1) - 2(x + h + 1)}{(x + h + 1)(x + 1)} = \frac{-2h}{(x + h + 1)(x + 1)}

f(x+h)βˆ’f(x)h=βˆ’2(x+h+1)(x+1)\frac{f(x+h) - f(x)}{h} = \frac{-2}{(x + h + 1)(x + 1)}

Answer: βˆ’2(x+h+1)(x+1)\frac{-2}{(x + h + 1)(x + 1)}.

Problem 5: Evaluate the piecewise function at x=βˆ’1x = -1, x=2x = 2, and x=5x = 5: f(x)={x2+1ifΒ x≀210βˆ’xifΒ x>2f(x) = \begin{cases} x^2 + 1 & \text{if } x \leq 2 \\ 10 - x & \text{if } x > 2 \end{cases}

f(βˆ’1)=(βˆ’1)2+1=2f(-1) = (-1)^2 + 1 = 2 (since βˆ’1≀2-1 \leq 2)

f(2)=(2)2+1=5f(2) = (2)^2 + 1 = 5 (since 2≀22 \leq 2)

f(5)=10βˆ’5=5f(5) = 10 - 5 = 5 (since 5>25 > 2)

Answer: f(βˆ’1)=2f(-1) = 2, f(2)=5f(2) = 5, f(5)=5f(5) = 5.

Key Takeaways

  • A function is formally a set of ordered pairs where each input maps to exactly one output
  • The domain is the set of valid inputs; the range is the set of actual outputs
  • Function notation f(x)f(x) names the output β€” parentheses do not mean multiplication
  • The difference quotient f(x+h)βˆ’f(x)h\frac{f(x+h)-f(x)}{h} measures average rate of change and is the gateway to calculus
  • Functions can be defined by formulas, tables, graphs, or verbal descriptions
  • The vertical line test determines whether a graph represents a function
  • Always substitute with care β€” use parentheses to protect against sign and order errors

Return to College Algebra for more topics in this section.

Last updated: March 29, 2026