Why Endpoints Need a One-Sided Definition
At an interior point of a function's domain, nearby inputs are available on both sides. At an endpoint, the domain may extend in only one direction. A difference quotient that uses an input outside the domain is not defined, so it cannot be used to calculate a derivative there.
For example, if the domain is \([0,\infty)\), then \(0\) is included but negative inputs are not. To study the change in \(f\) near \(0\), we must compare \(f(0)\) with values \(f(h)\) for positive \(h\), not negative \(h\). This leads to a one-sided derivative: a limit of difference quotients taken only through increments that remain in the domain.
Definition at the Endpoints of an Interval
Let \(f\) be defined on an interval \([a,b]\), where \(a<b\). At the left endpoint \(a\), the admissible nearby inputs are to the right, so the increment \(h\) must be positive. At the right endpoint \(b\), the admissible nearby inputs are to the left, so \(h\) must be negative.
The signs on \(h\) specify which increments are allowed: \(h>0\) at the left endpoint and \(h<0\) at the right endpoint. In both cases, \(h\) tends to zero through values for which the shifted input remains in the domain. If the endpoint is not in the domain, \(f\) has no value there to place in the quotient, so a derivative at that point is not defined.
A finite limit is required. If the quotient grows without bound or has no limit, the one-sided derivative does not exist as a finite real number. A one-sided derivative is therefore a derivative defined using one side; it is not a claim that a two-sided difference quotient is available at the endpoint.
| Point in the domain | Allowed increments | Endpoint quotient |
|---|---|---|
| Left endpoint \(a\) of \([a,b]\) | \(h>0\) | \(\frac{f(a+h)-f(a)}{h}\) |
| Right endpoint \(b\) of \([a,b]\) | \(h<0\) | \(\frac{f(b+h)-f(b)}{h}\) |
Worked Examples
Worked Example: A Polynomial at the Left Endpoint
Let \(f(x)=3x^2-2x+5\) on \([0,\infty)\). Find the derivative at the left endpoint \(0\).
The domain permits positive increments from \(0\), so we calculate the right-hand derivative. For \(h>0\),
As \(h\to 0^+\), \(3h-2\to -2\). Therefore \(f'_+(0)=-2\). Checking in the original quotient, the simplified quotient \(3h-2\) has limit \(-2\) through positive \(h\), as required.
Worked Example: A Polynomial at the Right Endpoint
Let \(g(x)=x+x^2\) on \([0,1]\). Find the derivative at the right endpoint \(1\).
At \(1\), nearby points in the domain lie to the left. Set \(h<0\), with \(1+h\in[0,1]\). Then
Taking \(h\to0^-\) gives \(3\), so \(g'_-(1)=3\). The quotient \(3+h\) approaches \(3\) for the allowed negative increments, confirming the endpoint derivative.
Worked Example: The Domain Determines the Relevant Side
Let \(q(x)=|x|\) on \([0,\infty)\). Find the derivative at \(0\).
For every admissible increment \(h>0\), \(|0+h|=h\), and \(q(0)=0\). Thus
The quotient is exactly \(1\) for every \(h>0\), so \(q'_+(0)=1\). The absolute-value formula has no negative inputs in this domain. Using negative increments to test this derivative would not describe the given function.
Worked Example: An Endpoint Quotient Without a Finite Limit
Let \(r(x)=\sqrt{x}\) on \([0,\infty)\). Does \(r\) have a finite derivative at \(0\)?
Only \(h>0\) is admissible. Since \(r(0)=0\),
As \(h\to0^+\), \(1/\sqrt{h}\) grows without bound. The quotient therefore has no finite real limit, so \(r'_+(0)\) does not exist as a finite derivative. This conclusion comes directly from the endpoint definition: the limit required there is not finite.
Why the Permitted Side Gives the Correct Derivative
A useful way to understand the one-sided definition is to compare a function on a closed interval with a differentiable function defined on a larger interval. The larger function has a two-sided derivative, but restricting its inputs to one side of an endpoint leaves the corresponding one-sided derivative unchanged.
For \(h>0\) sufficiently small, \(a+h\in[a,a+\delta]\), and the restriction agrees with \(F\) there. Thus \(\frac{f(a+h)-f(a)}{h}=\frac{F(a+h)-F(a)}{h}\).
The limit of the quotient for \(F\) as \(h\to0\) from both sides is \(L\). Consequently, its limit as \(h\to0^+\) is also \(L\).
The endpoint quotient for \(f\) equals the right-hand quotient for \(F\) for every sufficiently small positive \(h\). Therefore \(\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}=L\), so \(f'_+(a)=L\).
The same reasoning applies at a right endpoint: if \(F'(b)=L\) and \(f\) agrees with \(F\) on an interval immediately to the left of \(b\), then \(f'_-(b)=L\). The proof uses negative increments instead of positive ones. In either case, the endpoint quotient is simply the relevant one-sided part of the difference quotient for the larger function.
Common Mistake: Using an Increment Outside the Domain
Suppose \(f(x)=x^2\) has domain \([0,\infty)\), and we want the derivative at \(0\). A tempting but invalid calculation is to write the difference quotient for \(h<0\) and use \(f(h)=h^2\). This treats the formula \(x^2\) as if it defined the function at negative inputs, even though the stated function has no such inputs.
The problem is not the algebra: the problem is that \(f(h)\) is undefined for \(h<0\) in this example. Use \(h>0\) instead:
Thus \(f'_+(0)=0\). The formula \(x^2\) could also define a different function on all of \(\mathbb{R}\), but that would be a different domain. The derivative calculation must respect the domain of the function being studied.
What to Check at a Domain Endpoint
Before calculating, identify the endpoint and the inputs that actually belong to the domain near it. Then form the corresponding quotient and take its one-sided limit. A useful checklist is:
- Is the endpoint itself in the domain, so that the endpoint value is defined?
- Is the endpoint the left or right end of the interval?
- Which sign of \(h\) keeps the shifted input inside the domain?
- Does the quotient approach a finite real number through those admissible increments?
These checks distinguish three situations that should not be confused: a valid one-sided quotient with a finite limit, a valid quotient that has no finite limit, and an attempted quotient that is undefined because it uses inputs outside the domain. Only the first gives a finite endpoint derivative under the definition above.
Check Your Understanding
For each question, pay attention to the domain before forming a difference quotient.
- If \(f\) is defined on \([2,5]\), which sign of \(h\) is used to define its derivative at \(2\), and which sign at \(5\)?
- For \(p(x)=x^2-4x\) on \([0,\infty)\), set up the right-hand difference quotient at \(0\) and find its limit.
- For \(q(x)=3-x^2\) on \([1,4]\), set up the left-hand difference quotient at \(4\) and find its limit.
- Why is a negative increment inadmissible when differentiating a function whose domain is \([0,\infty)\) at \(0\)?
- What does it mean if an endpoint difference quotient is defined for all admissible increments but grows without bound as the increments approach zero?
