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Glacius
CalculusConcept reference

Corners and cusps

Differentiability fails when the one-sided difference quotients do not approach the same finite value.

1 min read1 figureSources & further reading
On this page 7 sections
  1. Overview
  2. For f(x)=|x|, both branches meet at (0,0)
  3. A cusp can have unbounded opposite slopes
  4. A derivative requires the same finite limiting slope from both sides
  5. Key takeaway
  6. Sources & further reading
  7. Concept connections

01For f(x)=|x|, both branches meet at (0,0)#

For f(x)=∣x∣f(x)=|x|, both branches meet at (0,0)(0,0). The graph is continuous there, but it has a corner.

Equal scale 30 pixels per unit: origin(160,145); y=|x| passes through(-3,3),(0,0),(3,3). The two branches meet continuously; the left slope is -1 and right slope +1, so no derivative exists at zero.Equal scale 30 pixels per unit: origin(160,145); y=|x| passes through(-3,3),(0,0),(3,3). The two branches meet continuously; the left slope is -1 and right slope +1, so no derivative exists at zero.
Figure 1Equal scale 30 pixels per unit: origin(160,145); y=|x| passes through(-3,3),(0,0),(3,3). The two branches meet continuously; the left slope is -1 and right slope +1, so no derivative exists at zero.
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For f(x)=∣x∣f(x)=|x| at 00, the quotient is ∣h∣/h|h|/h. Negative hh gives −1-1; positive hh gives 11.

lim⁡h→0−Q(h)=−1lim⁡h→0+Q(h)=1\begin{gathered}\lim_{h\to0^-}Q(h)=-1\\\lim_{h\to0^+}Q(h)=1\end{gathered}

The limits disagree. Their average is not a derivative.

Check your reasoning

Difference quotient: Q(h)=−2+2hQ(h)=-2+2h for h<0h<0; Q(h)=3−hQ(h)=3-h for h>0h>0. Derivative?

  1. AYes: average 1/21/2.
  2. BNo: −2≠3-2\ne3.
  3. CYes: right slope 33.
Show answer and explanation
No: −2≠3-2\ne3.

The side limits differ.

02A cusp can have unbounded opposite slopes#

A cusp can have unbounded opposite slopes. For f(x)=∣x∣f(x)=\sqrt{|x|} at 00, the quotient on the right is h/h=1/h\sqrt h/h=1/\sqrt h.

As h→0+h\to0^+, its positive denominator shrinks toward zero, so the quotient grows without bound.

For f(x)=∣x∣f(x)=\sqrt{|x|} at 00, negative hh gives Q(h)=−1/−hQ(h)=-1/\sqrt{-h}. Its positive denominator shrinks toward zero, so QQ decreases without bound.

lim⁡h→0−Q(h)=−∞lim⁡h→0+Q(h)=+∞\begin{gathered}\lim_{h\to0^-}Q(h)=-\infty\\\lim_{h\to0^+}Q(h)=+\infty\end{gathered}
Check your reasoning

At a response join, limiting secant rates are left 44 and right 44. One instantaneous rate?

  1. AYes: slope 44.
  2. BNo: two sides forbid it.
  3. CYes: slope 55.
Show answer and explanation
Yes: slope 44.

Both limits equal 44.

03A derivative requires the same finite limiting slope from both sides#

A derivative requires the same finite limiting slope from both sides. Unequal finite limits produce a corner. Opposite unbounded limits produce a cusp. Continuity alone does not decide differentiability.

Check your reasoning

Difference-quotient limits at 00: left −∞-\infty, right +∞+\infty. Claim: derivative 00. Repair.

  1. A00: signs cancel.
  2. B11: both sides steep.
  3. CNone: no finite limit.
Show answer and explanation
None: no finite limit.

No finite slope exists.

Key takeaway

Compare both limiting slopes; they must agree and be finite.

  • Diagnose failure of differentiability from unequal one-sided slopes.

Sources & further reading

  1. [1]

Reference this concept

Link to this page, a section, or an individual figure.

Glacius. “Corners and cusps.” Math behind ML. /learn/c-nondifferentiable