Why Endpoints Need a One-Sided Test
Square-root and fractional-power functions often have restricted domains. For example, \(f(x)=\sqrt{x}\) is defined for \(x\ge 0\), so \(x=0\) is an endpoint of its domain. At that point, values of the function are available to the right but not to the left. A difference quotient must respect this restriction.
The shape of the graph near an endpoint can be deceptive. A curve may approach the endpoint smoothly while its slopes grow without bound; another may have a horizontal tangent there. The difference quotient, with the domain restriction taken into account, distinguishes these cases.
Endpoint Derivatives and Their Domains
Suppose \(f\) is defined on an interval containing points to the right of \(a\), with \(a\) as its left endpoint. Its right-hand derivative at \(a\), when it exists, is the finite limit
If instead \(a\) is a right endpoint and the function is defined to its left, the left-hand derivative is
In both definitions, \(h\) must point into the domain. A limit of \(+\infty\) or \(-\infty\) is not a finite real limit, so it does not give a derivative. It indicates that the difference quotient is unbounded.
For an endpoint, this one-sided derivative is the relevant derivative test on the restricted domain. It does not assert that the function has a two-sided derivative at that point: a two-sided derivative requires the function to be defined on both sides nearby.
Fractional Powers at a Left Endpoint
Consider the basic family \(f(x)=c(x-a)^p+L\), defined for \(x\ge a\), where \(p>0\), \(c\ne 0\), and \(L\) is constant. Here the base \(x-a\) is nonnegative, so the fractional power is real-valued. At the endpoint \(a\), \(f(a)=L\). Put \(h=x-a\), so \(h>0\) for points in the domain immediately to the right of \(a\). Then
The exponent \(p-1\) determines the limit. If \(p<1\), then \(h^{p-1}=1/h^{1-p}\) grows without bound as \(h\to0^+\). If \(p=1\), the quotient is constant. If \(p>1\), then \(h^{p-1}\to0\).
This classifies the basic power family, but an added linear term can change a finite endpoint slope. For instance, adding \(mx\) adds \(m\) to the difference quotient; it does not remove an unbounded term when \(0<p<1\).
Why the Exponent Determines the Slope
The result follows directly from the definition, rather than from a power rule. The only limiting fact needed is that for \(q>0\), \(h^q\to0\) as \(h\to0^+\). For \(0<p<1\), the exponent \(1-p\) is positive, so \(h^{1-p}\to0^+\), and consequently its reciprocal grows without bound. For \(p>1\), the exponent \(p-1\) is positive, so \(h^{p-1}\to0\). When \(p=1\), that power is \(h^0=1\).
For \(f(x)=c(x-a)^p+L\), calculate \(f(a)=L\).
Because the domain is \(x\ge a\), write \(x=a+h\) with \(h>0\).
Subtract \(f(a)\) and divide by \(h\) to obtain \(c h^{p-1}\).
Compare \(p\) with \(1\), and check whether the resulting limit is finite.
Worked Examples
Worked Example: A Square-Root Endpoint with Unbounded Slope
Find the endpoint derivative, if it exists, for \(f(x)=\sqrt{x+2}\) at the left endpoint \(x=-2\). The domain condition \(x+2\ge0\) gives \(x\ge-2\), so use \(h>0\). First, \(f(-2)=\sqrt{0}=0\). The difference quotient is
As \(h\to0^+\), \(\sqrt{h}\to0^+\), so \(1/\sqrt{h}\to+\infty\). Therefore the endpoint derivative is not a finite real number. This agrees with the classification for \(p=\tfrac12<1\). Checking the quotient confirms that the positive, unbounded behavior comes from increments within the domain.
Worked Example: A Fractional Power with a Zero Endpoint Derivative
Find the endpoint derivative of \(g(x)=3(x-1)^{3/2}+2\) at \(x=1\), where \(x\ge1\). The endpoint value is \(g(1)=3(0)^{3/2}+2=2\). Let \(h>0\). Then
Since \(\sqrt{h}\to0\), the quotient tends to \(0\), so \(g'_+(1)=0\). This is a finite endpoint derivative. Substitution into the quotient verifies the result directly: its limiting value is \(3\cdot0=0\), as predicted for an exponent greater than one.
Worked Example: A Finite Nonzero Slope from a Linear Term
Find the endpoint derivative of \(q(x)=7-4x+2(x+1)^{3/2}\) at \(x=-1\), with domain \(x\ge-1\). First, \(q(-1)=7-4(-1)+2(0)^{3/2}=11\). For \(h>0\),
As \(h\to0^+\), this tends to \(-4+2(0)=-4\). Thus \(q'_+(-1)=-4\). The check is the difference quotient itself: the fractional-power contribution tends to zero, while the linear contribution remains \(-4\).
Right Endpoints: Powers of the Distance to the Endpoint
For a function defined to the left of \(a\), a useful form is \(f(x)=c(a-x)^p+L\), with \(x\le a\). Now \(f(a)=L\), and a valid increment satisfies \(h<0\). Since \(a-(a+h)=-h>0\), the quotient becomes
The same comparison with \(1\) applies to \(p\): when \(p>1\), the quotient tends to \(0\); when \(p=1\), it equals \(-c\); and when \(0<p<1\), its magnitude grows without bound. For the last case, the direction of the infinite limit depends on the sign of \(-c\).
Worked Example: A Square-Root Endpoint at the Right
Find the endpoint derivative, if it exists, for \(r(x)=5-2\sqrt{4-x}\), whose domain is \(x\le4\), at \(x=4\). The endpoint value is \(r(4)=5\). Use \(h<0\), so \(4-(4+h)=-h>0\). The quotient is
For the final equality, \(h=-(-h)\), with \(-h>0\). As \(h\to0^-\), \(-h\to0^+\), so \(2/\sqrt{-h}\to+\infty\). There is no finite endpoint derivative. The quotient is positive and unbounded, confirming both the sign and the classification for an exponent of \(\tfrac12\).
A Common Mistake: Using the Wrong Side
At \(x=0\), the function \(f(x)=\sqrt{x}\) is defined only for \(x\ge0\). A tempting but invalid calculation is to take \(h<0\) in the quotient and write
This is not a real-valued difference quotient: \(\sqrt{h}\) is not real for negative \(h\). The correction is to use \(h>0\), as the domain requires. Then
Another error is to treat an infinite limit as a derivative. The quotient for \(\sqrt{x}\) becomes unbounded, but a derivative must be a finite real limit. Therefore there is no finite endpoint derivative at \(0\), even though the quotient has a clear direction of growth.
Summary
For a power of the distance from an endpoint, the difference quotient reduces to a constant times a power whose exponent is \(p-1\). At a left endpoint, use positive increments; at a right endpoint, use negative increments. A power with \(0<p<1\), including a square root, produces an unbounded quotient. A power with \(p=1\) produces a finite constant slope, and a power with \(p>1\) contributes a zero endpoint slope. Added linear terms can supply a finite nonzero slope when the fractional-power contribution tends to zero.
Always check the domain before forming the quotient, then decide whether the limit is finite. This keeps the distinction clear between a one-sided endpoint derivative and a difference quotient that merely becomes unbounded.
Check Your Understanding
For each function, identify the valid direction for the endpoint quotient and decide whether the endpoint derivative is finite. Show the difference quotient that supports your answer.
- For \(f(x)=\sqrt{x-3}\) at \(x=3\), what is the limit of the endpoint difference quotient?
- For \(g(x)=(x+2)^{4/3}\) on \(x\ge-2\), what is the right-hand derivative at \(x=-2\)?
- For \(r(x)=\sqrt{5-x}\) at its right endpoint \(x=5\), which sign of \(h\) is valid, and does the quotient have a finite limit?
- For \(q(x)=6+3x+(x-1)^{3/2}\) on \(x\ge1\), find the endpoint derivative at \(x=1\).
- Why does a difference quotient tending to \(+\infty\) not count as an endpoint derivative?
