Domain and Range for MATH 139

Exam Relevance for MATH 139

Likelihood of appearing: Medium

Often tested as part of function analysis. Know domain restrictions from denominators, square roots, and logarithms.

Lesson

What is Domain and Range?

Every function is like a machine: you put something in (the input), and you get something out (the output). But not every machine accepts every input — and not every output is possible.

  • Domain = all the valid inputs (x-values) you can plug into the function
  • Range = all the possible outputs (y-values) the function can produce

Understanding domain and range helps you know where a function "lives" — which values make sense to use and which results you can expect.

Visualizing Domain and Range

Graph showing a parabola $f(x) = x^2$ with domain and range highlighted
Graph showing a parabola $f(x) = x^2$ with domain and range highlighted

In this graph, notice that:

  • Domain: You can plug in ANY x-value (the parabola extends left and right forever) → Domain: $(-\infty, \infty)$
  • Range: The outputs only go from 0 upward (the parabola never dips below the x-axis) → Range: $[0, \infty)$

Interval Notation Quick Reference

Notation Meaning When to Use
$(a, b)$ All numbers between $a$ and $b$, not including $a$ or $b$ Open interval
$[a, b]$ All numbers between $a$ and $b$, including $a$ and $b$ Closed interval
$[a, b)$ Includes $a$, excludes $b$ Half-open
$(-\infty, a)$ All numbers less than $a$ Unbounded left
$(a, \infty)$ All numbers greater than $a$ Unbounded right
$(-\infty, \infty)$ All real numbers No restrictions

💡 Key rule: Always use parentheses with $\infty$ or $-\infty$ — you can never "reach" infinity, so you can't include it!

Finding the Domain

The domain is usually "all real numbers" except where the function breaks. Look for these common restrictions:

Restriction 1: Division by Zero

You cannot divide by zero. If the function has a denominator, set it $\neq 0$.

$$f(x) = \frac{1}{x-3}$$

The denominator is $x - 3$. Set it $\neq 0$:

$$x - 3 \neq 0 \implies x \neq 3$$

Domain: $(-\infty, 3) \cup (3, \infty)$ — all real numbers except 3.

Restriction 2: Square Roots of Negatives

You cannot take the square root of a negative number (in real numbers). The expression under the root must be $\geq 0$.

$$g(x) = \sqrt{x + 2}$$

Set the inside $\geq 0$:

$$x + 2 \geq 0 \implies x \geq -2$$

Domain: $[-2, \infty)$

Restriction 3: Logarithms of Non-Positives

You cannot take the logarithm of zero or a negative number. The input to a log must be $> 0$.

$$h(x) = \ln(x - 5)$$

Set the inside $> 0$:

$$x - 5 > 0 \implies x > 5$$

Domain: $(5, \infty)$

Combining Restrictions

Some functions have multiple restrictions. Find each one, then combine.

$$f(x) = \frac{\sqrt{x}}{x - 4}$$

  • Square root: $x \geq 0$
  • Denominator: $x \neq 4$

Domain: $[0, 4) \cup (4, \infty)$

Finding the Range

The range is often trickier than the domain. Here are strategies:

Strategy 1: Use the Graph

Graph of $f(x) = \frac{1}{x}$ showing horizontal asymptote
Graph of $f(x) = \frac{1}{x}$ showing horizontal asymptote

For $f(x) = \frac{1}{x}$:

  • The graph approaches but never touches $y = 0$
  • The function produces every other y-value
  • Range: $(-\infty, 0) \cup (0, \infty)$ — all real numbers except 0

Strategy 2: Think About What Outputs Are Possible

For $f(x) = x^2$:

  • Squaring any number gives a non-negative result
  • You can get 0 (when $x = 0$)
  • You can get arbitrarily large values (as $x \to \pm\infty$)
  • Range: $[0, \infty)$

Strategy 3: Solve for x in Terms of y

Set $y = f(x)$ and solve for $x$. The values of $y$ that give valid solutions are in the range.

For $f(x) = 2x + 1$:

$$y = 2x + 1$$ $$y - 1 = 2x$$ $$x = \frac{y - 1}{2}$$

This gives a valid $x$ for ANY value of $y$. Range: $(-\infty, \infty)$

Example 1: Complete Domain and Range Analysis

Find the domain and range of $f(x) = \sqrt{9 - x^2}$.

Step 1: Find the domain

The expression under the square root must be $\geq 0$:

$$9 - x^2 \geq 0$$ $$9 \geq x^2$$ $$-3 \leq x \leq 3$$

Domain: $[-3, 3]$

Step 2: Find the range

  • The minimum value of $\sqrt{9 - x^2}$ is $0$ (when $x = \pm 3$)
  • The maximum value is $\sqrt{9} = 3$ (when $x = 0$)

Range: $[0, 3]$

Step 3: Visualize

This is actually the top half of a circle with radius 3!

Example 2: Rational Function

Find the domain and range of $f(x) = \frac{x + 1}{x - 2}$.

Step 1: Find the domain

Denominator $\neq 0$:

$$x - 2 \neq 0 \implies x \neq 2$$

Domain: $(-\infty, 2) \cup (2, \infty)$

Step 2: Find the range

Solve for $x$:

$$y = \frac{x + 1}{x - 2}$$

$$y(x - 2) = x + 1$$

$$yx - 2y = x + 1$$

$$yx - x = 2y + 1$$

$$x(y - 1) = 2y + 1$$

$$x = \frac{2y + 1}{y - 1}$$

This is undefined when $y = 1$. So $y = 1$ is NOT in the range.

Range: $(-\infty, 1) \cup (1, \infty)$

Example 3: Piecewise Function from a Graph

Using the graph below, find the domain and range of the piecewise function.

Graph of a piecewise function with closed/open dots
Graph of a piecewise function with closed/open dots

When reading domain and range from a graph:

  • Domain: Look at how far left and right the graph extends (x-axis coverage)
  • Range: Look at how far up and down the graph extends (y-axis coverage)
  • Closed dots (●) mean the point IS included — use a bracket $[$
  • Open dots (○) mean the point is NOT included — use a parenthesis $($

Finding the Domain:

  • The first piece extends from the upper left (toward $x = -\infty$) and ends at $x = 0$ with an open circle
  • The second piece starts at $x = 0$ with a closed dot and ends at $x = 3.5$ with an open circle
  • The third piece starts at $x = 3.5$ with an open circle and extends to the lower right (toward $x = +\infty$)

At $x = 0$: The closed dot means $x = 0$ IS included.

At $x = 3.5$: Both adjacent pieces have open circles, so $x = 3.5$ is NOT included.

$$\boxed{\text{Domain: } (-\infty, 3.5) \cup (3.5, \infty)}$$

Finding the Range:

  • The first piece has an arrow pointing up toward $+\infty$
  • The third piece has an arrow pointing down toward $-\infty$
  • Together, every y-value from $-\infty$ to $+\infty$ is covered

$$\boxed{\text{Range: } (-\infty, \infty)}$$

Common Mistakes and Misunderstandings

❌ Mistake: Using brackets with infinity

Wrong: $[0, \infty]$

Why it's wrong: Infinity is not a number — you can never "reach" it. You can't include something that doesn't exist as a specific value.

Correct: $[0, \infty)$


❌ Mistake: Forgetting to check ALL restrictions

Wrong: For $f(x) = \frac{\sqrt{x}}{x-1}$, saying domain is $x \neq 1$

Why it's wrong: You found the denominator restriction but forgot the square root restriction ($x \geq 0$).

Correct: Domain: $[0, 1) \cup (1, \infty)$


❌ Mistake: Confusing domain and range

Wrong: "The range of $f(x) = x^2$ is all real numbers because you can plug in any x."

Why it's wrong: That describes the domain, not the range. The range asks what outputs are possible.

Correct: Domain: $(-\infty, \infty)$. Range: $[0, \infty)$ because $x^2$ is never negative.

Formulas & Reference

Domain

$$\text{Domain} = \{x : f(x) \text{ is defined}\}$$

The set of all valid input values (x-values) for which the function produces a real output.

Variables:
$x$:
input value
$f(x)$:
the function

Range

$$\text{Range} = \{y : y = f(x) \text{ for some } x \text{ in domain}\}$$

The set of all possible output values (y-values) that the function can produce.

Variables:
$y$:
output value
$f(x)$:
the function
$x$:
input from the domain

Square Root Domain Restriction

$$\sqrt{\text{expression}} \implies \text{expression} \geq 0$$

For real numbers, you cannot take the square root of a negative. Set the expression under the root ≥ 0 and solve.

Variables:
$\text{expression}$:
whatever is under the square root

Denominator Restriction

$$\frac{\text{numerator}}{\text{denominator}} \implies \text{denominator} \neq 0$$

Division by zero is undefined. Set the denominator ≠ 0 and solve to find excluded values.

Variables:
$\text{denominator}$:
the bottom of the fraction

Logarithm Domain Restriction

$$\ln(\text{expression}) \implies \text{expression} > 0$$

Logarithms are only defined for positive inputs. Set the expression inside the log > 0 and solve.

Variables:
$\text{expression}$:
the input to the logarithm
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