Differentiability: Corners, Cusps & More

A function is differentiable at x = a when its derivative f'(a) exists, which means the limit of the difference quotient is a finite number. Differentiability fails at corners, cusps, vertical tangents and discontinuities. Every differentiable function is continuous, but a continuous function can still fail to be differentiable, as |x| does at 0.

Before this lesson: Definition of the Derivative, Sketching the Graph of a Derivative, Continuity and One-Sided Limits

What differentiable means

Zoom in on a smooth curve and it starts to look like a straight line, its tangent line. A function is differentiable at a point when this happens: there is one tangent line with a finite slope.

Some graphs never straighten out, no matter how far you zoom. At a sharp corner, the left side and the right side keep their different slopes. At a cusp or a vertical tangent, the slope grows without bound. At a break, there is no single curve to zoom in on. In each case, f'(a) does not exist.

On a graph, differentiable means smooth and unbroken: no corners, no sharp points, no vertical tangent lines and no gaps.

Three small graphs that are continuous at x = 0 but not differentiable there. Corner: y = |x| makes a V shape whose slopes jump from −1 to 1. Cusp: y = x^(2/3) comes to a sharp point where the slopes go to negative infinity on the left and positive infinity on the right. Vertical tangent: y = x^(1/3) passes through the origin with a vertical tangent line, so the slope is infinite. — All three graphs are continuous at 0, but none has a derivative there: a corner, a cusp and a vertical tangent.

When a derivative does not exist

Situation at x = a	Why f'(a) fails	Example
Corner	The left and right slopes are different numbers	$\|x\|$ at 0
Cusp	The slopes go to $-\infty$ on one side and $\infty$ on the other	$x^{2/3}$ at 0
Vertical tangent	The slope goes to $\infty$ (or $-\infty$ ) on both sides	$x^{1/3}$ at 0
Discontinuity	A hole, jump or asymptote; f is not continuous	$\frac{1}{x}$ at 0

Differentiability implies continuity: if f is differentiable at a, then f is continuous at a. The reverse is false; |x| is continuous at 0 but not differentiable.

Piecewise functions: f is differentiable at a boundary point a when (1) f is continuous at a, and (2) the slopes of the two pieces match at a.

Differentiability examples

Example 1: A corner

Show that f(x) = |x| is not differentiable at x = 0.

\frac{f(0 + h) - f(0)}{h} = \frac{|h|}{h}

Write the difference quotient at 0.

\lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1

For h less than 0, |h| = −h. This is the slope from the left.

\lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1

For h greater than 0, |h| = h. This is the slope from the right.

-1 \ne 1

The one-sided limits differ (see one-sided limits).

Answer: the limit does not exist, so f'(0) does not exist. The graph has a corner at the origin, even though |x| is continuous there.

Example 2: A vertical tangent

Is $f(x) = x^{1/3}$ differentiable at x = 0?

\frac{f(h) - f(0)}{h} = \frac{h^{1/3}}{h}

Write the difference quotient at 0.

\frac{h^{1/3}}{h} = \frac{1}{h^{2/3}}

Subtract exponents: $\frac{1}{3} - 1 = -\frac{2}{3}$ (see rational exponents).

h^{2/3} \gt 0

$h^{2/3} = (h^{1/3})^2$ is a positive number near 0, on both sides.

\lim_{h \to 0} \frac{1}{h^{2/3}} = \infty

Positive over a tiny positive number.

Answer: no. The slope grows without bound, so the graph has a vertical tangent line at the origin and f'(0) does not exist.

Example 3: A cusp

Is $f(x) = x^{2/3}$ differentiable at x = 0?

\frac{f(h) - f(0)}{h} = \frac{h^{2/3}}{h} = \frac{1}{h^{1/3}}

Write the difference quotient and subtract exponents.

\lim_{h \to 0^-} \frac{1}{h^{1/3}} = -\infty

For h less than 0, the cube root is negative.

\lim_{h \to 0^+} \frac{1}{h^{1/3}} = \infty

For h greater than 0, it is positive.

Answer: no. The slopes blow up in opposite directions, which makes a sharp point called a cusp.

Example 4: A piecewise function that is smooth

Let $f(x) = x^2$ for $x \le 1$ and $f(x) = 2x - 1$ for $x \gt 1$ . Is f differentiable at x = 1?

1^2 = 1

Continuity: the left piece gives f(1) = 1.

2(1) - 1 = 1

The right piece approaches 1, so f is continuous at 1.

2(1) = 2

Slope from the left: the derivative of $x^2$ is 2x (see definition of the derivative), which is 2 at x = 1.

2 = 2

Slope from the right: the line 2x − 1 has slope 2. The slopes match.

Answer: yes, f is differentiable at 1, with f'(1) = 2. The line is the tangent line to the parabola at (1, 1), so the pieces join smoothly.

Example 5: Word problem with a skateboard ramp

A skateboard ramp follows the curve $y = 0.1x^2$ for x from 0 to 5 meters, then continues as a straight line y = ax + b. Find a and b so the ramp has no bump or corner at x = 5.

0.1(5)^2 = 2.5

The curve’s height at x = 5.

0.2(5) = 1

The curve’s slope at x = 5: the derivative of $0.1x^2$ is 0.2x.

a = 1

For no corner, the line’s slope must match.

5a + b = 2.5

For no gap, the line must have the same height at x = 5.

5 + b = 2.5

Substitute a = 1.

b = -2.5

Subtract 5 from both sides.

Answer: a = 1 and b = −2.5, so the straight part is y = x − 2.5. Matching height makes the ramp continuous, and matching slope makes it differentiable, so a skateboard rolls across the join smoothly.

Common mistakes

Mistake	Example of the error	Fix
Thinking continuous means differentiable	“\|x\| is continuous at 0, so it has a derivative”	Continuity is needed but not enough. Check that the slopes agree.
Checking only the slopes of the pieces	Calling f differentiable because both pieces have slope 3 at the boundary, while there is a jump	Check continuity first. A jump means no derivative.
Calling a vertical tangent a zero slope	Saying $x^{1/3}$ has f'(0) = 0	The tangent is vertical, so the slope is undefined (infinite).
Using the wrong piece for a side	Using 2x − 1 for the left-hand slope in Example 4	Left of 1 uses $x^2$ ; right of 1 uses 2x − 1.
Thinking a function can’t be differentiable at a piece boundary	Assuming every piecewise function has a corner	Pieces that match in height and slope join smoothly.
Mixing up cusp and corner	Calling \|x\| a cusp	Corner: finite slopes that differ. Cusp: slopes go to $\pm\infty$ .

Related methods

Why differentiable implies continuous

If f'(a) exists, then $f(a + h) - f(a) = \frac{f(a + h) - f(a)}{h} \cdot h$ , which approaches $f'(a) \cdot 0 = 0$ . So f(a + h) approaches f(a), which is continuity.

Derivatives on a graph

On the graph of f’, a corner of f becomes a jump, and a vertical tangent of f becomes a vertical asymptote. See sketching the graph of a derivative.

Practice problems

Is f(x) = |x − 3| differentiable at x = 3? Explain.
Is $f(x) = x^2 + 1$ differentiable everywhere? Explain.
Is $f(x) = (x - 2)^{1/3}$ differentiable at x = 2?
Is $f(x) = (x + 1)^{2/3}$ differentiable at x = −1?
Is $f(x) = \frac{1}{x}$ differentiable at x = 0?
Let f(x) = 3x − 2 for $x \lt 1$ and $f(x) = x^2$ for $x \ge 1$ . Is f differentiable at x = 1?
Let $f(x) = x^2$ for $x \lt 2$ and f(x) = 4x − 4 for $x \ge 2$ . Is f differentiable at x = 2?
Find a and b so that $f(x) = x^2 + 1$ for $x \le 2$ and f(x) = ax + b for $x \gt 2$ is differentiable at x = 2.
True or false: if f is continuous at a, then f is differentiable at a.
Shipping costs C(w) = 5 + 2w dollars for w ≤ 10 pounds and C(w) = 25 + 1.5(w − 10) for w > 10. Is C differentiable at w = 10? What does that mean?

Show answers

\lim_{h \to 0^-} \frac{|h|}{h} = -1

The difference quotient at 3 is $\frac{|h|}{h}$ ; from the left it is −1.

\lim_{h \to 0^+} \frac{|h|}{h} = 1

From the right it is 1.

Answer: no. The slopes −1 and 1 differ, so the graph has a corner at x = 3.

f'(x) = 2x

The derivative exists for every x.

Answer: yes. Polynomials are smooth everywhere, with no corners, cusps, vertical tangents or breaks.

\frac{h^{1/3}}{h} = \frac{1}{h^{2/3}} \to \infty

The difference quotient at 2 grows without bound on both sides.

Answer: no; there is a vertical tangent at x = 2.

\frac{h^{2/3}}{h} = \frac{1}{h^{1/3}}

The difference quotient at −1.

Answer: no. It goes to $-\infty$ from the left and $\infty$ from the right: a cusp.

f(0) \text{ is undefined}

f is not continuous at 0.

Answer: no. A function must be continuous to be differentiable.

3(1) - 2 = 1 = 1^2

Continuity: both pieces give 1 at x = 1.

3 \ne 2(1)

Slopes: the line has slope 3, and $x^2$ has slope 2x = 2 at x = 1.

Answer: no. f is continuous at 1, but the slopes differ, so there is a corner.

2^2 = 4 = 4(2) - 4

Continuity: both pieces give 4.

2(2) = 4

Slopes: $x^2$ has slope 2x = 4 at x = 2, and the line has slope 4.

Answer: yes, and f'(2) = 4.

a = 2(2) = 4

Match the slopes: $x^2 + 1$ has slope 2x, which is 4 at x = 2.

2a + b = 5

Match the heights: $2^2 + 1 = 5$ .

8 + b = 5

Substitute a = 4.

b = -3

Subtract 8 from both sides.

Answer: a = 4 and b = −3

f(x) = |x|

A counterexample: |x| is continuous at 0 but not differentiable there.

Answer: false.

10.

5 + 2(10) = 25

Continuity: the first rule gives C(10) = 25, and the second rule also approaches 25.

2 \ne 1.5

Slopes: $2 per pound on the left, $1.50 per pound on the right.

Answer: no. C is continuous at 10 pounds but has a corner there: the cost per extra pound drops from $2 to $1.50.

Differentiability FAQ

What does it mean for a function to be differentiable?

It means the derivative exists: the limit of the difference quotient is a finite number. On a graph, the curve is smooth and has a single, non-vertical tangent line.

Where is a function not differentiable?

At corners, cusps, vertical tangent lines and discontinuities (holes, jumps and asymptotes).

Does differentiability imply continuity?

Yes. If f'(a) exists, f is continuous at a. The reverse is not true: |x| is continuous at 0 but not differentiable there.

How do you check if a piecewise function is differentiable?

At the boundary, check that the pieces meet (continuity) and that their slopes are equal. If both hold, the function is differentiable there.

Summary

f is differentiable at a when f'(a) exists as a finite number.
Derivatives fail at corners, cusps, vertical tangents and discontinuities.
Differentiable implies continuous, but continuous does not imply differentiable.
For piecewise functions, match both the heights and the slopes at the boundary.
On a graph, differentiable means smooth: no sharp points, no vertical tangents and no breaks.

Previous lesson: Sketching the Graph of a Derivative

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