Why Endpoints Need a Separate Derivative Test
At an interior point of a function’s domain, nearby inputs are available on both sides. The usual derivative limit therefore compares the function’s change to the left and to the right. At an endpoint of a restricted domain, one of those sides is missing: the function may only be defined for inputs to the right of a left endpoint, or to the left of a right endpoint.
The derivative idea still makes sense there, provided we use only inputs that belong to the domain. We take a one-sided difference-quotient limit from inside the domain. This is called an endpoint derivative or a one-sided derivative at an endpoint.
Definitions on a Closed Interval
Suppose \(f\) is defined on an interval \([a,b]\), where \(a<b\). The left endpoint \(a\) has domain points immediately to its right, and the right endpoint \(b\) has domain points immediately to its left.
“Finite” matters: a limit that grows without bound, or fails to settle to a single real number, does not define an endpoint derivative. The formulas use the endpoint value \(f(a)\) or \(f(b)\), so the function must be defined at the endpoint as well as on the inward side.
These definitions extend to other restricted domains when the domain contains an interval immediately to the inward side of the endpoint. They do not provide a derivative at a point merely because it is called a boundary point: there must be domain inputs approaching from the relevant side.
Why the Inward Difference Quotient Has This Form
The endpoint formulas are the ordinary difference quotient with the input restricted to the available side. To see the direction and sign explicitly, set \(x=a+h\) at a left endpoint, where \(h>0\). Then \(x\) approaches \(a\) from the right, and \(x-a=h\). At a right endpoint, set \(x=b-h\), again with \(h>0\). Now \(x\) approaches \(b\) from the left, and \(x-b=-h\).
Substitute \(x=a+h\), with \(h>0\), into the difference quotient. The denominator is \(x-a=h\), so the right-hand derivative is \(\displaystyle\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}\).
Substitute \(x=b-h\), with \(h>0\). The denominator is \(x-b=-h\), so the left-hand derivative is \(\displaystyle\lim_{h\to0^+}\frac{f(b-h)-f(b)}{-h}\).
At the right endpoint, \(x-b\) is negative for inward inputs \(x<b\). Replacing it by \(h\) without also changing the numerator’s orientation changes the sign of the quotient.
This substitution gives equivalent forms of the definitions, not a new derivative rule. It is useful because the positive increment \(h\) makes clear which direction the input is approaching from.
Endpoint Derivatives Versus Interior Derivatives
At an interior point \(c\), the function has domain inputs on both sides in a sufficiently small interval around \(c\). The ordinary derivative exists if the two-sided limit \(\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\) exists as a finite real number. Equivalently, the left-hand and right-hand difference-quotient limits must both exist and have the same value.
At the left endpoint of \([a,b]\), only the right-hand limit is required by the endpoint convention. At the right endpoint, only the left-hand limit is required. There is no two-sided derivative test at an endpoint of the restricted domain, because the quotient cannot be formed from both sides using inputs in the domain.
| Location | Inputs used | Derivative test |
|---|---|---|
| Interior point \(c\) | Inputs to the left and right of \(c\) | A finite two-sided difference-quotient limit |
| Left endpoint \(a\) | Inputs to the right of \(a\) | A finite right-hand difference-quotient limit |
| Right endpoint \(b\) | Inputs to the left of \(b\) | A finite left-hand difference-quotient limit |
Some definitions reserve “differentiable” for points where a two-sided derivative is available. In the common closed-interval convention, one says a function is differentiable on \([a,b]\) when it has the ordinary derivative at interior points and the appropriate one-sided derivatives at the endpoints. The important point is to state which convention is being used. In either convention, the endpoint calculation itself uses the inward one-sided limit.
Worked Endpoint Calculations
Worked Example: Derivatives at Both Ends of a Restricted Interval
Let \(f(x)=x^2+3x+1\) on \([0,2]\). Calculate its endpoint derivatives.
At the left endpoint, \(f(0)=1\). For \(h>0\),
Therefore,
Check: the expression \(h+3\) is exactly the original inward difference quotient after simplification, and it approaches \(3\) as the allowed increment approaches \(0\).
At the right endpoint, \(f(2)=4+6+1=11\). Write an inward input as \(2-h\), with \(h>0\). Then
The denominator for this input is \((2-h)-2=-h\), so
Thus \(f'_-(2)=\displaystyle\lim_{h\to0^+}(7-h)=7\). Check: the inward quotient simplifies to \(7-h\), whose limit is \(7\); the negative denominator at this endpoint has been retained.
Worked Example: A Rational Function at a Left Endpoint
Let \(g(x)=\dfrac{1}{1+x}\) on \([0,2]\). Find the right-hand derivative at \(x=0\) directly from the definition.
First, \(g(0)=1\). For \(h>0\),
Taking the inward limit gives
Check: substituting the simplified quotient back into the defining expression preserves its value for every \(h>0\), and that expression approaches \(-1\). The endpoint derivative is therefore finite and equals \(-1\).
Worked Example: A Quotient Without a Finite Endpoint Derivative
Define \(q\) on \([0,1]\) by \(q(0)=0\) and \(q(x)=1\) for \(x>0\). Does \(q\) have a right-hand derivative at \(0\)?
For every \(h>0\), \(q(h)=1\), so the endpoint difference quotient is
As \(h\to0^+\), \(1/h\) increases without bound. It does not approach a finite real number, so \(q'_+(0)\) does not exist as an endpoint derivative.
Check: for any positive number \(M\), choosing \(0<h<1/M\) gives \(1/h>M\). Thus the quotient cannot approach a finite value, confirming that the endpoint derivative does not exist.
A Common Mistake: Demanding a Two-Sided Limit
For \(f(x)=x^2\) restricted to \([0,1]\), consider the left endpoint \(0\). A common incorrect conclusion is: “The derivative does not exist because there are no domain inputs less than \(0\), so a two-sided limit cannot be taken.” That applies the interior-point test where the endpoint test is needed.
The correct calculation uses the available side. Since \(f(0)=0\), for \(h>0\),
Therefore the right-hand endpoint derivative is \(f'_+(0)=0\). The conclusion is not that a two-sided derivative has been evaluated on \([0,1]\). It is that the inward one-sided derivative exists. If the function were instead given on a larger domain, a two-sided derivative would be a separate test based on that larger domain.
What to Check in an Endpoint Problem
Before calculating, identify the endpoint and the side that remains inside the domain. Then use the corresponding endpoint value and keep the denominator \(x-a\) or \(x-b\) consistent with the chosen input. Finally, check that the quotient approaches a finite real number. A one-sided limit that diverges or oscillates is not an endpoint derivative.
Endpoint derivatives provide a precise way to discuss instantaneous change at the boundary of a restricted domain. They also prevent a common confusion: a function can have a well-defined inward rate of change at an endpoint even though the two-sided derivative test cannot be applied there. At interior points, the two-sided criterion remains essential; at endpoints, the domain determines the appropriate one-sided test.
Check Your Understanding
For each question, identify the relevant one-sided limit before evaluating it.
- For \(f(x)=x^2-2x\) on \([0,3]\), write the difference quotient for the right-hand derivative at \(0\).
- For the same function, write the inward difference quotient at the right endpoint \(3\), using \(h>0\).
- If an endpoint quotient approaches \(-4\) from the inward side, what is the endpoint derivative?
- Why is a two-sided derivative test not required at the left endpoint of a restricted interval?
- Does a quotient that increases without bound as the inward increment approaches \(0\) define a finite endpoint derivative? Explain.
