Find the real domain of $f(x)=\sqrt{7-2x}$.
- Require $7-2x\ge0$.
- $x\le7/2$, so the domain is $(-\infty,7/2]$.
Why is the square-root endpoint included?
Zero is allowed inside a square root, so the endpoint belongs to the domain.
Functions, domain and range
Separate allowed inputs from the outputs a function actually reaches.
By the end of this lesson
- use function notation and distinguish a function from a general relation
- find domains from restrictions
- find ranges from rules, domains and graphs
Understand the idea
Under the rule $f(x)=x^2$, inputs −2 and 2 both produce 4. That is allowed: a function gives each input exactly one output, but different inputs may share an output.
The domain tells us which inputs may enter. The range tells us which outputs are reached after applying the rule to those inputs. Changing the allowed inputs can change the range even when the formula stays the same.
Interactive guide
Check the logic
For y=x² with domain [1,2], what is the minimum output?
Explore the result
The vertex at x=0 is excluded. The smallest allowed input is 1, giving y=1.
Move the lower bound from −1 to 1. Explain when the minimum stops being zero.
For y=x² on a≤x≤2, a=−2 gives range [0,4]. If a becomes positive, zero is no longer attained and the range is [a²,4].
Put the idea into notation
For $f:D\to\mathbb R$, the range is $\{f(x):x\in D\}$. The target set written after the arrow need not be the actual range.
Over the reals, denominators cannot be zero and square-root arguments must be nonnegative. Logarithm arguments, when present, must be strictly positive. Intersect all restrictions.
On a graph, project onto the horizontal axis to read the domain and onto the vertical axis to read the range. A vertical line may meet a function graph at most once. Open endpoints are excluded; closed endpoints are included.
Checkpoint
Correct. The uniqueness condition goes from input to output.
Not quite. That stronger condition is one-to-one, not the definition of a function. Input signs do not determine whether a relation is a function.
Show explanation
The uniqueness condition goes from input to output.
Example 1
Apply the input restriction
Example 2
Let the domain determine the range
Find the range of $g(x)=(x-1)^2+2$ for $-1\le x\le3$.
- The minimum occurs at $x=1$, giving 2.
- The endpoints both give 6; every intermediate output is reached.
- The range is $[2,6]$.
Why check the vertex as well as endpoints?
A domain interval need not map to endpoint outputs in the same order. Include the vertex when it lies inside the interval.
Example 3
Keep the hole after simplifying
Find the domain and range of $h(x)=(x^2-1)/(x-1)$.
- The original denominator excludes $x=1$. For every other input, $h(x)=x+1$.
- The domain is $\mathbb R\setminus\{1\}$. The missing input would have produced 2.
- No other input of $x+1$ produces 2, so the range is $\mathbb R\setminus\{2\}$.
Why does the hole remain after cancellation?
Cancelling a factor simplifies the rule on its original domain; it does not put an excluded input back. Here that missing input also removes one output.
In the exam
Quick check
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The range describes reached outputs.
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The range describes reached outputs.
Substitute 4 into the rule: 12 − 2 = 10.
Show explanation
Substitute 4 into the rule: 12 − 2 = 10.
The root allows x = 0, but placing it in the denominator excludes zero; together the conditions require x > 0.
Show explanation
The root allows x = 0, but placing it in the denominator excludes zero; together the conditions require x > 0.
The minimum is at 0 and the maximum is 4 at −2.
Show explanation
The minimum is at 0 and the maximum is 4 at −2.
The smallest allowed input is 2, giving output 4; squaring increases from there.
Show explanation
The smallest allowed input is 2, giving output 4; squaring increases from there.
Lesson Rewards
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