When Slopes Become Unbounded in Opposite Directions
A graph can be continuous at a point and still fail to have a derivative there. At a corner, the one-sided slopes approach different finite values. At a cusp, the one-sided difference quotients instead grow without bound, with opposite signs. The graph comes to a sharp point, and its secant slopes become increasingly steep as they approach that point from either side.
The key is to examine the difference quotient separately on the left and on the right. A graph that merely looks pointed is not enough to establish cusp behavior: the limits of those quotients determine what happens.
The Difference Quotients to Compare
For \(h\ne 0\), the difference quotient at \(a\) is
It is the slope of the secant line through the graph points at \(x=a\) and \(x=a+h\). To test for a cusp, calculate its limit as \(h\to 0^-\) and as \(h\to 0^+\), keeping the signs of \(h\) in view.
The derivative \(f'(a)\) exists only if the two-sided difference quotient has a finite real limit. If the left and right quotients instead tend to opposite infinities, there is no such finite limit. We describe the behavior of the slopes as unbounded; we do not assign the derivative a value of infinity.
Worked Example: A Basic Fractional-Power Cusp
Worked Example: \(f(x)=|x|^{2/3}\) at the origin
First, \(f(0)=0\). For \(h\ne 0\), the difference quotient is
If \(h>0\), then \(h=|h|\), so
As \(h\to 0^+\), the positive denominator approaches zero, so this quotient tends to \(+\infty\). For example, at \(h=1/1000\), it equals \(1/(1/10)=10\).
If \(h<0\), then \(h=-|h|\), so
As \(h\to 0^-\), this quotient tends to \(-\infty\). For example, at \(h=-1/1000\), it equals \(-10\). The function is continuous at the origin because \(|x|^{2/3}\to 0=f(0)\), and the one-sided quotients have opposite infinite limits. Thus the graph has a cusp at \(x=0\), and \(f'(0)\) does not exist as a finite real number.
Check: Substituting \(h=1/1000\) and \(h=-1/1000\) into the original quotient gives \(10\) and \(-10\), respectively, matching the simplified expressions.
Why the Exponent Matters
For the family \(f(x)=f(a)+c|x-a|^p\), where \(c\ne 0\) and \(p>0\), set \(h=x-a\). The difference quotient becomes
Its magnitude is \(|c||h|^{p-1}\). When \(0<p<1\), the exponent \(p-1\) is negative, so this magnitude grows without bound as \(|h|\to 0\). The sign comes from both the coefficient \(c\) and the sign of \(h\). This explains why absolute-value fractional powers with exponent between zero and one are common sources of cusps.
Why This Result Holds
The calculation above shows the shape of the quotient. The following argument makes the unbounded limits precise and verifies that the signs oppose one another.
As \(x\to a\), \(|x-a|^p\to 0\) because \(p>0\). Hence \(f(x)\to f(a)\).
Put \(h=x-a\). For \(h>0\), the quotient is \(c h^{p-1}\). For \(h<0\), it is \(-c|h|^{p-1}\).
Since \(0<p<1\), \(1-p>0\) and \(|h|^{p-1}=1/|h|^{1-p}\). Given any \(M>0\), choose \(\delta=(|c|/M)^{1/(1-p)}\). Whenever \(0<|h|<\delta\), the quotient's magnitude is \(|c|/|h|^{1-p}>M\).
The right quotient has the sign of \(c\), while the left quotient has the opposite sign. Their limits are therefore infinite with opposite signs.
These steps establish the result: continuity holds at \(a\), and arbitrarily close to \(a\), the two one-sided quotients exceed any prescribed magnitude while pointing in opposite directions. Since a finite derivative would require a finite two-sided limit, it cannot exist at the cusp.
Worked Example: A Shifted and Scaled Cusp
Worked Example: \(g(x)=3-2|x-1|^{1/2}\) at \(x=1\)
At the point \(a=1\), \(g(1)=3\). For \(h\ne 0\),
For \(h>0\), this is \(-2/h^{1/2}\), which tends to \(-\infty\) as \(h\to0^+\). For \(h<0\), using \(h=-|h|\), it is
which tends to \(+\infty\) as \(h\to0^-\). For a numerical check, if \(h=1/100\), the quotient is \(-2/(1/10)=-20\). If \(h=-1/100\), it is \(2/(1/10)=20\). Also, \(g(x)\to3=g(1)\) as \(x\to1\), so \(g\) is continuous there. The opposing infinite limits confirm a cusp at \(x=1\), not a finite derivative.
Check: Substitution in the original quotient gives \(-2\sqrt{1/100}/(1/100)=-20\) on the right and \(-2\sqrt{1/100}/(-1/100)=20\) on the left, as found above.
Worked Example: Unequal Fractional Powers on the Two Sides
Worked Example: A cusp with different steepness on each side
Define \(q(0)=0\), and let
Both pieces approach zero as \(x\to0\), so \(q\) is continuous at zero. For \(h<0\), the quotient is
which tends to \(-\infty\). For \(h>0\), it is
which tends to \(+\infty\). For \(h=-1/8\), the left quotient is \(-2/(1/8)^{2/3}=-2/(1/4)=-8\). For \(h=1/8\), the right quotient is \(5/(1/8)^{2/3}=5/(1/4)=20\). The magnitudes differ, but the signs oppose, so the point is a cusp and the derivative at zero does not exist as a finite real number.
Check: Directly, \(q(-1/8)=2(1/2)=1\), so \(q(-1/8)/(-1/8)=-8\). Also \(q(1/8)=5(1/2)=5/2\), so \(q(1/8)/(1/8)=20\).
A Common Sign Mistake
A frequent error is to replace \(h\) by \(|h|\) in the denominator of the difference quotient before considering whether \(h\) is negative. That changes the quotient and can erase the very sign needed to identify a cusp.
Worked Example: Correcting a lost sign
For \(f(x)=|x|^{2/3}\), an incorrect calculation for \(h<0\) is
The mistake is that \(h=-|h|\), not \(h=|h|\), when \(h<0\). The correct calculation is
For \(h=-1/8\), the correct quotient is \(-1/(1/8)^{1/3}=-2\), not \(2\). Keeping the sign of the increment gives the left-hand limit \(-\infty\), opposite the right-hand limit \(+\infty\).
What to Look For
When a graph or formula suggests a cusp, use a short checklist rather than relying on appearance alone:
- Is the function defined at the point, and does it approach its value there?
- What is the difference quotient for increments \(h<0\)?
- What is the difference quotient for increments \(h>0\)?
- Do the quotient magnitudes grow without bound as \(h\to0\)?
- Do the two limits have opposite signs?
All of these features matter. A sharp-looking graph does not by itself establish a cusp, and unbounded slopes alone do not establish the opposing-sign behavior that defines one here. The difference quotient test supplies the evidence: continuity at the point and opposite infinite one-sided limits.
Check Your Understanding
For each question, keep the sign of the increment explicit when working with a left-hand quotient.
- For \(f(x)=|x|^{1/2}\), write the difference quotient at zero separately for \(h<0\) and \(h>0\). What are its one-sided limits?
- For \(g(x)=4+3|x-2|^{1/3}\), determine the signs of the left- and right-hand quotient limits at \(x=2\).
- Why does a pair of opposite infinite one-sided limits show that the derivative does not exist as a finite real number?
- In the family \(f(x)=f(a)+c|x-a|^p\), what restriction on \(p\) makes the quotient magnitude unbounded near \(a\)?
- A learner replaces \(h\) with \(|h|\) in a quotient while evaluating the left-hand limit. What information may this incorrectly remove?
