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how to find domain and range

To find the domain and range of a function, think in two stages:

  1. what x-values are allowed to go in, and
  2. what y-values actually come out.

What “domain” and “range” mean

  • Domain : all input values xxx for which the formula makes sense (the function is defined).
  • Range : all output values y=f(x)y=f(x)y=f(x) that you actually get when you plug in domain values.

In set-builder form, you often see
Domain(f)={x∈R:condition on x}\text{Domain}(f)=\{x\in \mathbb{R}:\text{condition on }x\}Domain(f)={x∈R:condition on x} and
Range(f)={f(x):x∈Domain(f)}\text{Range}(f)=\{f(x):x\in \text{Domain}(f)\}Range(f)={f(x):x∈Domain(f)}.

Step-by-step: how to find the domain (formula-based)

When you have a formula y=f(x)y=f(x)y=f(x), you try to find all real xxx that do not break the rules. Common “danger zones”:

  1. Denominators
    • You cannot divide by 0.
    • Rule: Solve “denominator ≠0\neq 0=0” and exclude those x-values.
    • Example: f(x)=14x−3f(x)=\frac{1}{4x-3}f(x)=4x−31​.
      • Denominator 4x−3≠0⇒x≠344x-3\neq 0\Rightarrow x\neq \frac{3}{4}4x−3=0⇒x=43​.
   * Domain: all real numbers except x=34x=\frac{3}{4}x=43​.
  1. Square roots (and other even roots)
    • You cannot take the square root of a negative real number (in basic real-valued functions).
    • Rule: expression under the root ≥0\ge 0≥0.
    • Example: f(x)=x+2f(x)=\sqrt{x+2}f(x)=x+2​.
      • Need x+2≥0⇒x≥−2x+2\ge 0\Rightarrow x\ge -2x+2≥0⇒x≥−2.
   * Domain: [−2,∞)[-2,\infty)[−2,∞).
  1. Logarithms
    • Inside of log⁡(⋅)\log(\cdot)log(⋅) must be positive.
    • Rule: argument >0>0>0.
    • Example: f(x)=log⁡(x−1)f(x)=\log(x-1)f(x)=log(x−1).
      • Need x−1>0⇒x>1x-1>0\Rightarrow x>1x−1>0⇒x>1.
      • Domain: (1,∞)(1,\infty)(1,∞).
  2. Piecewise functions

    • The domain is the union of all intervals stated in the pieces.

    • Example (made-up):
      f(x)=x2f(x)=x^2f(x)=x2 if x<0x<0x<0; f(x)=x+2f(x)=x+2f(x)=x+2 if x≥0x\ge 0x≥0.

      • Domain: all real numbers (because pieces cover all xxx).
  1. Polynomials and simple exponentials
    • Polynomials like x2+1x^2+1x2+1, exponentials like 2x2^x2x usually have domain = all real numbers.

Step-by-step: how to find the range (formula-based)

Range can be trickier. A powerful general strategy is:

  1. Write y=f(x)y=f(x)y=f(x).
  2. Solve for xxx in terms of yyy: get x=g(y)x=g(y)x=g(y).
  3. Ask: for which yyy values does x=g(y)x=g(y)x=g(y) make sense? Those yyy form the range of fff.

Example (rational function):

  • y=14x−3y=\dfrac{1}{4x-3}y=4x−31​.
  • Solve for xxx:
    • y(4x−3)=1⇒4yx−3y=1y(4x-3)=1\Rightarrow 4yx-3y=1y(4x−3)=1⇒4yx−3y=1.
    • 4yx=1+3y⇒x=1+3y4y4yx=1+3y\Rightarrow x=\dfrac{1+3y}{4y}4yx=1+3y⇒x=4y1+3y​.
  • Now, for what yyy is this defined?
    • Denominator 4y≠0⇒y≠04y\neq 0\Rightarrow y\neq 04y=0⇒y=0.
  • So the range is all real numbers except 000.

Example (quadratic):

  • f(x)=x2+1f(x)=x^2+1f(x)=x2+1.
  • Domain: all real numbers.
  • Minimum value of x2x^2x2 is 0, so minimum of x2+1x^2+1x2+1 is 1; no upper bound.
  • Range: [1,∞)[1,\infty)[1,∞).

Example (square root type):

  • General square-root function: f(x)=ax+bf(x)=\sqrt{ax+b}f(x)=ax+b​ with a≠0a\ne 0a=0.
  • Domain is all xxx with ax+b≥0ax+b\ge 0ax+b≥0, which is a half-line.
  • Square root outputs are always ≥0\ge 0≥0, so range is [0,∞)[0,\infty)[0,∞).

How to find domain and range from a graph

If you have a graph:

  1. Domain from graph
    • Look left to right along the x-axis.
    • Ask: for which x-values does the graph actually have points?
    • Include endpoints if the graph has a solid dot or closed shape there; exclude if there is an open circle.
  1. Range from graph
    • Look bottom to top along the y-axis.
    • Ask: what y-values are covered by the graph?
    • Again, use open/closed points or arrows to decide whether endpoints are included.

Example (visual idea):

  • If the graph starts at x=−2x=-2x=−2 and goes right forever, domain is [−2,∞)[-2,\infty)[−2,∞).
  • If the lowest point on the graph is y=−3y=-3y=−3 and it extends upward without bound, range is [−3,∞)[-3,\infty)[−3,∞).

Typical function types and their domain/range

Here is a compact reference:

html

<table>
  <tr>
    <th>Function type</th>
    <th>Example</th>
    <th>Domain (real-valued)</th>
    <th>Range (real-valued)</th>
  </tr>
  <tr>
    <td>Linear</td>
    <td>f(x) = 2x + 1</td>
    <td>All real numbers</td>
    <td>All real numbers</td>
  </tr>
  <tr>
    <td>Quadratic (upward)</td>
    <td>f(x) = x^2 + 1</td>
    <td>All real numbers</td>
    <td>[1, ∞)</td>
  </tr>
  <tr>
    <td>Square root</td>
    <td>f(x) = √(x + 2)</td>
    <td>x ≥ -2</td>
    <td>[0, ∞)</td>
  </tr>
  <tr>
    <td>Rational (simple)</td>
    <td>f(x) = 1 / (4x − 3)</td>
    <td>All x ≠ 3/4</td>
    <td>All y ≠ 0</td>
  </tr>
  <tr>
    <td>Logarithmic</td>
    <td>f(x) = ln(x − 1)</td>
    <td>x &gt; 1</td>
    <td>All real numbers</td>
  </tr>
</table>

The rational example and the quadratic example align with standard textbook treatments and worked examples.

Quick “algorithm” you can memorize

When you see a new function f(x)f(x)f(x):

  1. Start by assuming domain is all real numbers.
  2. Remove any x that
    • makes a denominator 0,
    • makes an even root negative,
    • makes a log argument ≤ 0,
    • violates any explicit condition (like “x≥0x\ge 0x≥0”).
  1. To find range, either
    • reason from the shape (quadratic, square root, etc.), or
    • solve y=f(x)y=f(x)y=f(x) for xxx and find when that new expression is defined.
  1. If you have a graph , scan horizontally for domain and vertically for range.

Information gathered from public forums or data available on the internet and portrayed here.