When a Graph Has a Corner
A curve can be continuous and still fail to have a derivative at a point. One common reason is a corner: the graph reaches the point from both sides, but its slope changes abruptly there. The graph of \(f(x)=|x|\) at \(x=0\) is a familiar example. Its two straight branches meet, but they approach the meeting point with different slopes.
To test for a corner, compare the slopes calculated from the left and from the right. If both slopes approach finite values but those values are unequal, there is no single slope that describes the graph at the point. The derivative therefore does not exist there.
Continuity matters in this definition: the graph must meet at the point. The focus here is the change in slope at such a meeting, not every possible reason a derivative might fail to exist.
One-Sided Slopes
The derivative at \(x=a\) is based on the difference quotient \(\frac{f(a+h)-f(a)}{h}\) as \(h\) approaches zero. For a left-hand derivative, \(h\) approaches zero through negative values; for a right-hand derivative, \(h\) approaches zero through positive values.
These limits, when finite, give the slopes the graph approaches from each side. The two-sided derivative exists only if the left-hand and right-hand limits are equal and finite. In particular, unequal finite one-sided slopes cannot be combined into one derivative.
Why Unequal Slopes Rule Out a Derivative
Let \(Q(h)=\frac{f(a+h)-f(a)}{h}\) be the difference quotient at \(a\). The ordinary derivative \(f'(a)\), if it exists, is the two-sided limit of \(Q(h)\) as \(h\) approaches zero. A two-sided limit must have the same value whether \(h\) approaches zero through positive or negative values. Therefore, the left-hand and right-hand limits of \(Q(h)\) must agree.
Use the same difference quotient on both sides: \(Q(h)=\frac{f(a+h)-f(a)}{h}\).
Calculate the limit as \(h\to0^-\) and as \(h\to0^+\).
If the finite limits differ, \(Q(h)\) has no two-sided limit, so \(f'(a)\) does not exist.
This argument is a direct consequence of the limit definition: if a two-sided limit existed, both one-sided limits would have to equal it. Thus unequal one-sided slopes do more than suggest a sharp-looking graph; they prove that the derivative is absent at that point.
Worked Example: The Corner in \(f(x)=|x|\)
Worked Example: The Corner in \(f(x)=|x|\)
Find the one-sided derivatives of \(f(x)=|x|\) at \(a=0\), and decide whether \(f'(0)\) exists.
First, \(f(0)=|0|=0\). For \(h\ne0\), the difference quotient is
If \(h>0\), then \(|h|=h\), so the quotient is \(\frac{h}{h}=1\). Therefore \(f'_+(0)=1\).
If \(h<0\), then \(|h|=-h\), so the quotient is \(\frac{-h}{h}=-1\). Therefore \(f'_-(0)=-1\).
The function is continuous at zero because \(\lim_{x\to0}|x|=0=f(0)\). Its two finite one-sided slopes are different: \(-1\ne1\). Hence \(f'(0)\) does not exist, and the graph has a corner at \(x=0\).
Check: Substituting any positive \(h\) into the quotient gives \(1\); substituting any negative \(h\) gives \(-1\). These values confirm the two one-sided slopes.
Worked Example: A Shifted and Scaled Absolute Value
Worked Example: A Shifted and Scaled Absolute Value
Consider \(g(x)=|2x-3|+1\). Find and compare the one-sided slopes at \(a=\frac32\).
At the point, \(g\left(\frac32\right)=|3-3|+1=1\). Write a nearby input as \(x=\frac32+h\). Then
The difference quotient is therefore
For \(h>0\), \(|2h|=2h\), so the quotient is \(2\), giving \(g'_+\left(\frac32\right)=2\). For \(h<0\), \(|2h|=-2h\), so the quotient is \(-2\), giving \(g'_-\left(\frac32\right)=-2\).
The function is continuous there: as \(h\to0\), \(g\left(\frac32+h\right)=|2h|+1\to1=g\left(\frac32\right)\). Since \(-2\ne2\), the derivative does not exist at \(x=\frac32\); this is a corner.
Check: The quotient equals \(2\) for every positive \(h\) and \(-2\) for every negative \(h\), confirming the computed one-sided slopes.
Worked Example: Two Linear Pieces Meeting
Worked Example: Two Linear Pieces Meeting
Let
Determine whether the graph has a corner at \(x=1\).
Since the first formula includes \(x=1\), \(f(1)=2(1)+1=3\). The value approached from the right is also \(-1+4=3\), so the two pieces meet at the point \((1,3)\). More explicitly, the left-hand values approach \(3\) because \(2x+1\to3\) as \(x\to1^-\), and the right-hand values approach \(3\) because \(-x+4\to3\) as \(x\to1^+\). Thus \(f\) is continuous at \(1\).
For \(h<0\), \(1+h\le1\), so the left-hand difference quotient is
Hence \(f'_-(1)=2\). For \(h>0\), \(1+h>1\), so the right-hand quotient is
Hence \(f'_+(1)=-1\). Both slopes are finite, but \(2\ne-1\), so \(f'(1)\) does not exist. The graph has a corner at the join.
Check: The left-hand expression gives the point value \(3\) at \(x=1\), and the right-hand expression also gives \(3\) there. The difference quotients simplify to the slopes of the corresponding linear pieces, \(2\) and \(-1\).
A Common Mistake: Choosing One Side’s Slope
For \(f(x)=|x|\), a mistaken argument might say: “To the right of zero, the graph is \(y=x\), which has slope \(1\). Therefore \(f'(0)=1\).” This uses only the right-hand behavior. The derivative at an interior point, however, requires the difference quotient to approach the same value from both sides.
The correction is to report the two one-sided slopes separately, then compare them. Since they disagree, neither one is the two-sided derivative. In particular, assigning a value to a formula for the slope at \(x=0\) cannot make the derivative exist there; the difference quotient determines whether it exists.
What the Slope Comparison Tells You
At a corner, the graph has a well-defined meeting point and a finite direction of approach on each side, but those directions do not agree. The derivative represents a single instantaneous slope, so two different one-sided slopes cannot describe one derivative value.
| At the point | What to compare | Conclusion for a corner |
|---|---|---|
| The function’s values | Does the graph meet continuously? | Yes |
| Left-hand behavior | Is the limiting slope finite? | Yes |
| Right-hand behavior | Is the limiting slope finite? | Yes |
| One-sided slopes | Are their values unequal? | Yes; the two-sided derivative does not exist |
For a reliable test, calculate from the difference quotient rather than relying only on how sharp the graph appears. A graph can help you anticipate the answer, but the one-sided limits establish it. Also keep the conclusion specific: unequal finite one-sided slopes prove a corner at that point; they do not mean that every failure of differentiability is a corner.
Check Your Understanding
For each question, show enough reasoning to identify the relevant one-sided slopes.
- For \(f(x)=|x|\), what are the left- and right-hand derivatives at zero, and why is there no derivative there?
- For \(g(x)=|x-2|+4\), write the difference quotient at \(x=2\) using \(h\), then find its two one-sided limits.
- In the piecewise example, why is it necessary to check that the function is continuous at \(x=1\) as well as compare the slopes?
- A continuous graph has a left-hand derivative of \(-3\) and a right-hand derivative of \(-3\) at a point. Does the corner criterion apply? Explain what the one-sided comparison tells you.
- A student finds a right-hand slope of \(5\) at a point and concludes that the derivative is \(5\). What additional calculation is needed?
