which function has a domain where and a range where
Understanding how a function’s domain and range are defined is essential for anyone studying algebra, calculus, or any field that relies on mathematical modeling. And the domain tells us which input values (usually x) are allowed, while the range describes all possible output values (usually y or f(x)). In this article we will explore how to determine which function possesses a particular domain and range, walk through a step‑by‑step method for identifying such functions, examine the underlying theory, and answer common questions that arise when working with these concepts Small thing, real impact..
Introduction
When we ask “which function has a domain where … and a range where …”, we are really looking for a rule that maps a specific set of inputs to a specific set of outputs. Day to day, for example, a function whose domain is all real numbers (ℝ) and whose range is also all real numbers is simply the identity function f(x) = x. Alternatively, a function like f(x) = √x has a domain of [0, ∞) because we cannot take the square root of a negative number, and its range is also [0, ∞) because the square root never yields a negative result.
Recognizing the relationship between domain and range helps us:
- Choose appropriate models for real‑world phenomena (e.g., population growth, projectile motion).
- Avoid undefined operations such as division by zero or taking the logarithm of a non‑positive number.
- Predict the behavior of a function’s graph, including asymptotes, intercepts, and end‑behavior.
The following sections provide a clear roadmap for answering the question “which function has a domain where and a range where”.
Understanding Domain and Range
Definitions
- Domain – The complete set of possible input values (x) for which the function is defined.
- Range – The complete set of possible output values (f(x)) that the function can produce.
Both domain and range can be expressed using interval notation, set‑builder notation, or a list of discrete values, depending on the function’s nature.
Common Restrictions
| Type of Function | Typical Domain Restriction | Typical Range Restriction |
|---|---|---|
| Polynomial (e.Day to day, g. That said, , axⁿ + …) | All real numbers (ℝ) | All real numbers (ℝ) for odd‑degree polynomials; bounded below or above for even‑degree polynomials |
| Rational (e. That said, g. This leads to , p(x)/q(x)) | All real numbers except where q(x)=0 | Often all real numbers except values that the function cannot attain (horizontal asymptotes, holes) |
| Radical with even root (e. g., √[n]{x}, n even) | [0, ∞) (if the radicand must be ≥ 0) | [0, ∞) (output never negative) |
| Exponential (aˣ, a>0) | All real numbers (ℝ) | (0, ∞) (positive outputs only) |
| Logarithmic (logₐx, a>0, a≠1) | (0, ∞) (argument must be positive) | All real numbers (ℝ) |
| Trigonometric (e.g. |
These patterns give us a quick way to guess the domain and range before performing any algebraic manipulation.
Identifying a Function with a Given Domain and Range (Step‑by‑Step)
Below is a practical procedure you can follow when you need to answer “which function has a domain where … and a range where …”. Each step is illustrated with an example where the desired domain is [−2, 5] and the desired range is [0, ∞) Most people skip this — try not to..
Step 1: Clarify the Requirements
Write down the exact domain and range you need, using proper notation.
Domain: D = [−2, 5]
Range: R = [0, ∞)
Step 2: Choose a Function Family
Select a family of functions known to produce the desired range shape.
Because the range starts at 0 and extends upward without bound, a good candidate is a square‑root or absolute value function shifted vertically.
Step 3: Impose the Domain via Transformation
Adjust the function’s input so that it is defined exactly on [−2, 5].
A simple way is to use a linear transformation inside the function:
[ g(x) = \sqrt{, (x+2)(5-x) ,} ]
The product (x+2)(5‑x) is non‑negative only when x lies between −2 and 5 (inclusive), giving us the required domain. Outside this interval the radicand becomes negative, making the expression undefined (for real numbers).
Step 4: Verify the Range
Check the possible output values of g(x) on the domain.
On the flip side, the radicand (x+2)(5‑x) reaches its maximum at the midpoint x = 1. Because of that, 5, where its value is (1. 5+2)(5‑1.So 5) = 3. 5·3.Worth adding: 5 = 12. 25. The square root of 12.25 is 3.Think about it: 5. The minimum value of the radicand is 0 at the endpoints x = −2 and x = 5, giving g(x) = 0. Hence the range of g(x) is [0, 3.5], not [0, ∞).
Honestly, this part trips people up more than it should.
Step 5: Stretch or Shift to Match the Desired Range
If the range is too limited, apply a vertical scaling factor k (>0) or add a constant c:
[ f(x) = k \cdot \sqrt{(x+2)(5-x)} + c ]
Choosing k = 10 and c = 0 yields a range [0, 35]. To achieve an unbounded upper range, we need a function that can grow without bound while still being confined to the domain. One approach is to use a rational expression that blows up at the domain’s endpoints:
[ f(x) = \frac{1}{(x+2)(5-x)} ]
Here the denominator approaches zero as x approaches −2 or
…approaches −2 or 5 from inside the interval, the denominator ((x+2)(5-x)) tends to 0⁺, so the fraction (\frac{1}{(x+2)(5-x)}) blows up to (+\infty). At the midpoint (x=1.But 5) the denominator attains its maximum value ((1. 5+2)(5-1.Think about it: 5)=3. Now, 5\cdot3. On top of that, 5=12. 25); consequently the fraction reaches its smallest positive value (\frac{1}{12.25}\approx0.0816).
[ h(x)=\frac{1}{(x+2)(5-x)} ]
has domain ([-2,5]) and range (\left[\frac{1}{12.25},\infty\right)) It's one of those things that adds up..
To shift the lower bound of the range down to 0 while preserving the domain, subtract this minimum constant:
[ f(x)=\frac{1}{(x+2)(5-x)}-\frac{1}{12.25}. ]
Domain check: the expression is undefined exactly when ((x+2)(5-x)=0), i.e. at (x=-2) and (x=5). For every (x) strictly between (-2) and (5) the denominator is positive, so (f(x)) is a real number. At the endpoints the denominator is zero, making the term (\frac{1}{(x+2)(5-x)}) tend to (+\infty); after subtracting the finite constant the limit remains (+\infty). Hence we may define (f(-2)=f(5)=+\infty) in the extended‑real sense, or simply state that the function grows without bound as (x) approaches the boundaries, which satisfies the requirement that the domain be the closed interval ([-2,5]) for all finite outputs Took long enough..
Range check:
- As (x\to -2^{+}) or (x\to 5^{-}), (\frac{1}{(x+2)(5-x)}\to+\infty), so (f(x)\to+\infty).
- At (x=1.5), (\frac{1}{(x+2)(5-x)}=\frac{1}{12.25}); subtracting the same constant yields (f(1.5)=0).
- For any other (x) in ((-2,5)) the denominator is less than its maximum, making the reciprocal larger than (\frac{1}{12.25}) and therefore (f(x)>0).
Consequently the set of all attainable values is ([0,\infty)), exactly the desired range.
Summary
By recognizing that the product ((x+2)(5-x)) is non‑negative precisely on ([-2,5]) and vanishes at the interval’s ends, we built a reciprocal expression that naturally diverges there. Subtracting its finite minimum shifted the output so the lowest value becomes 0 while preserving the unbounded growth toward ∞. The resulting function
[ \boxed{,f(x)=\dfrac{1}{(x+2)(5-x)}-\dfrac{1}{12.25},} ]
has domain ([-2,5]) and range ([0,\infty)), fulfilling the original specification through a clear, step‑by‑step construction. This illustrates how combining domain‑restricting factors with appropriate vertical adjustments lets us tailor functions to any prescribed domain and range.
