When a Point Has Two Different Slopes
A graph can be continuous at a point and still fail to have a derivative there. One common reason is a corner: the graph approaches the point with one finite slope from the left and a different finite slope from the right.
To diagnose this situation, calculate two limits separately. The left-hand derivative describes the rate of change as the input approaches the point from smaller values; the right-hand derivative describes the rate from larger values. If both limits exist and are finite but unequal, there is a corner, not a single derivative.
It is important to distinguish this from a discontinuity. A jump or other break in the function is not a corner merely because the graph changes direction nearby. The one-sided difference quotients are based on the actual value \(f(a)\), and a mismatch between that value and the nearby function values can prevent a finite one-sided slope.
One-Sided Derivatives
Let \(f\) be defined on an open interval containing \(a\). For \(h\ne 0\), the difference quotient at \(a\) is \(\frac{f(a+h)-f(a)}{h}\). The left- and right-hand derivatives are the corresponding one-sided limits, when they exist as finite real numbers.
Both quotients use the same value \(f(a)\). Only the side from which \(h\) approaches zero changes. This matters especially for piecewise-defined functions: the formula on each side may differ, but the numerator always subtracts the function’s value at the point itself.
When both one-sided derivatives exist and are finite, the one-sided derivative test says that the ordinary two-sided derivative exists exactly when they are equal. Thus, unequal finite one-sided derivatives rule out a derivative at \(a\).
What Counts as a Corner?
Continuity is part of this definition because a corner is a meeting of two finite-slope pieces at a shared point. If the function jumps instead, the graph does not meet there continuously; that is a different kind of failure. We will see below why a discontinuity also prevents finite one-sided derivatives defined using \(f(a)\).
Geometrically, the slopes of the two pieces describe the directions in which the graph arrives at and leaves the point. Equal slopes allow a single slope to describe both sides. Unequal finite slopes produce a sharp change in direction.
Worked Example: The Absolute-Value Corner
Worked Example: Finding the Slopes of \(f(x)=|x|\) at Zero
Let \(f(x)=|x|\) and \(a=0\). First, \(f(0)=|0|=0\). For \(h<0\), \(|h|=-h\), so the left difference quotient is
It follows that \(f'_-(0)=\lim_{h\to0^-}(-1)=-1\). For \(h>0\), \(|h|=h\), and therefore
Thus \(f'_+(0)=\lim_{h\to0^+}1=1\). The function is continuous at zero because \(\lim_{x\to0}|x|=0=f(0)\). Its one-sided derivatives are finite but unequal, so it has a corner at zero and is not differentiable there.
Check: The left-hand formula for the function is \(-x\), whose slope is \(-1\); the right-hand formula is \(x\), whose slope is \(1\). These agree with the difference-quotient limits.
Worked Example: A Piecewise Join with Unequal Slopes
Worked Example: Testing a Join at \(x=1\)
Consider the function
At the join, the first branch gives \(f(1)=1^2=1\). For \(h<0\), the input \(1+h\) is less than \(1\), so use the quadratic branch:
Taking \(h\to0^-\) gives \(f'_-(1)=2\). For \(h>0\), use the linear branch. Since \(f(1+h)=3(1+h)-2=1+3h\),
Taking \(h\to0^+\) gives \(f'_+(1)=3\). The function is continuous at the join: the left-hand values approach \(1\), the right-hand values approach \(3(1)-2=1\), and \(f(1)=1\). The one-sided slopes are finite and unequal, so \(x=1\) is a corner.
Check: The derivative of \(x^2\) at \(1\) is \(2(1)=2\), and the slope of \(3x-2\) is \(3\). These agree with the direct one-sided computations.
Why a Discontinuity Cannot Be a Finite-Slope Corner
The difference quotient itself explains why continuity is necessary for a corner. Suppose a one-sided derivative at \(a\) exists as a finite number \(L\). On that same side, write the change in function value as
As \(h\) approaches zero from that side, the quotient approaches \(L\), so it remains bounded near zero. The factor \(h\) approaches zero. Their product therefore approaches zero, which means \(f(a+h)\) approaches \(f(a)\) from that side. Thus a finite one-sided derivative based at \(f(a)\) requires the function to approach its value from that side. If there is a jump on that side, the one-sided derivative cannot be finite.
For a corner, both one-sided derivatives are finite. Applying this reasoning from each side shows that the function approaches \(f(a)\) from both sides. The function is consequently continuous at \(a\). A discontinuity and a corner, as defined here, are distinct diagnoses.
Worked Example: A Jump, Not a Corner
Worked Example: A Discontinuous Join at Zero
Define
At zero, the second branch gives \(f(0)=0+1=1\). As \(x\to0^-\), the first branch approaches \(0\), while as \(x\to0^+\), the second branch approaches \(1\). The two one-sided limits differ, so the function is discontinuous at zero.
Now calculate the difference quotients. If \(h<0\), then \(f(h)=h\), and
As \(h\to0^-\), \(\frac{1}{h}\to-\infty\), so \(1-\frac{1}{h}\to+\infty\). There is no finite left-hand derivative. If \(h>0\), then \(f(h)=h+1\), giving
The right-hand derivative is \(1\), but having one finite one-sided derivative is not enough for a corner. Since the function is discontinuous and the left-hand quotient has no finite limit, zero is not a corner.
Check: The left branch approaches \(0\), not \(f(0)=1\); this nonzero difference in function values is the constant \(-1\) in the left numerator. Dividing that term by \(h\) produces the unbounded term \(-1/h\), as the calculation shows.
A Common Mistake: Using One Formula on Both Sides
For \(f(x)=|x|\), a tempting but incorrect calculation is to write \(|h|=h\) and conclude that the difference quotient at zero is \(h/h=1\). This identity applies only when \(h>0\). When \(h<0\), the correct identity is \(|h|=-h\), so the quotient is \((-h)/h=-1\). The two sides must be evaluated under their respective sign conditions; using a single branch for both sides hides the corner.
Another error is to call any point where the derivative fails to exist a corner. The one-sided limits must be checked: a corner requires continuity and two finite, unequal one-sided derivatives. A discontinuity, an infinite one-sided slope, or a failure of a one-sided limit is not classified as a corner by this definition.
A Reliable Diagnostic
Determine \(f(a)\) from the definition of the function, including which branch applies at the join.
Use the formula valid for \(h<0\), and simplify \(\frac{f(a+h)-f(a)}{h}\) before taking the limit.
Use the formula valid for \(h>0\), keeping the same point value \(f(a)\) in the numerator.
If both limits are finite and unequal, check that the function is continuous at \(a\); then it is a corner. If the function is discontinuous, it is not a corner.
This method separates three questions that are easy to confuse: whether the function is continuous at the point, whether each one-sided slope exists as a finite number, and whether those slopes agree. A corner occurs only when the function is continuous, both slopes are finite, and the slopes differ.
Check Your Understanding
For each question, use the one-sided difference quotient where needed and state the conditions supporting your conclusion.
- For \(f(x)=|x-2|\), calculate the left- and right-hand derivatives at \(x=2\). Is the point a corner?
- For \(g(x)=\begin{cases}2x+1,&x\le 0,\\-x+1,&x>0,\end{cases}\) find \(g'_-(0)\) and \(g'_+(0)\), and classify the point.
- Why must a corner, under the definition in this tutorial, be a point of continuity?
- For \(p(x)=\begin{cases}x,&x<1,\\x+2,&x\ge1,\end{cases}\) identify \(p(1)\) and the two one-sided limits of the function at \(1\). Is \(1\) a corner?
- In the calculation for \(|h|/h\), which expression applies when \(h<0\), and why can the \(h>0\) expression not be used on the left?
