Differentiability and Continuity

A function is differentiable at a point when it has a derivative there, which means its graph has a single, non-vertical tangent line. Most functions you meet are differentiable almost everywhere, but not quite everywhere. This page shows exactly where derivatives break down, and how differentiability connects to continuity.

Key ideas

What differentiable means

$f$ is differentiable at $x = a$ if the limit

$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

exists as a finite number. Like any limit, it exists only if the left-hand limit ($h \to 0^-$) and the right-hand limit ($h \to 0^+$) are equal. These one-sided limits are the slopes from the left and from the right.

A function is differentiable on an interval if it is differentiable at every point of the interval.

Differentiable implies continuous (but not the other way)

• If $f$ is differentiable at $x = a$, then $f$ is continuous at $x = a$.
• The reverse is false: a function can be continuous at a point and still not be differentiable there. The classic example is $f(x) = |x|$ at $x = 0$.

Flip the first statement around and you get a useful test: if $f$ is not continuous at $a$, then $f$ is not differentiable at $a$.

Four ways a derivative can fail to exist

Type	What the graph looks like	What goes wrong
Corner	two straight-ish pieces meet at an angle	the slopes from the left and right are different numbers
Cusp	a sharp point	the slopes go to $+\infty$ on one side and $-\infty$ on the other
Vertical tangent	the graph is momentarily vertical	the slopes go to $+\infty$ (or $-\infty$) from both sides
Discontinuity	a hole, jump, or asymptote	not continuous, so not differentiable

A corner, a cusp, and a vertical tangent. All three functions are continuous at $x = 0$, but none is differentiable there.

On the AP exam you usually don’t need the name. You need the reason: ”$f$ is not differentiable at $x = 2$ because the slopes from the left and right are not equal” or “because $f$ is not continuous at $x = 2$”.

Checking a piecewise function

For a piecewise function made of “nice” pieces (like polynomials), check the boundary point $x = a$ in two steps:

1. Continuity: the two pieces must give the same value at $x = a$ (and that must be $f(a)$).
2. Matching slopes: the derivatives of the two pieces must give the same value at $x = a$.

If both hold, $f$ is differentiable at $a$. If step 1 fails, stop: $f$ is not differentiable, no matter what the slopes do.

For step 2 you need derivatives of the pieces. From the definition, $\dfrac{d}{dx}(mx + b) = m$, $\dfrac{d}{dx}\big(x^2\big) = 2x$, and $\dfrac{d}{dx}\big(x^3\big) = 3x^2$. The power rule makes these quick.

Worked examples

Example 1: A corner

Show that $f(x) = |x - 2|$ is continuous but not differentiable at $x = 2$.

Solution. $f(2) = 0$ and $\displaystyle\lim_{x \to 2} |x - 2| = 0$, so $f$ is continuous at $2$.

The difference quotient at $a = 2$ is $\dfrac{|2 + h - 2| - 0}{h} = \dfrac{|h|}{h}$.

• From the right ($h \gt 0$): $\dfrac{|h|}{h} = \dfrac{h}{h} = 1$, so the limit is $1$.
• From the left ($h \lt 0$): $\dfrac{|h|}{h} = \dfrac{-h}{h} = -1$, so the limit is $-1$.

The one-sided limits are different, so $f'(2)$ does not exist. The graph has a corner at $(2, 0)$.

Example 2: A vertical tangent and a cusp

Use difference quotients at $x = 0$ to explain the shapes of $y = \sqrt[3]{x}$ and $y = \sqrt[3]{x^2}$ in the figure.

Solution. For $f(x) = \sqrt[3]{x} = x^{1/3}$:

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

Since $h^{2/3} = \big(\sqrt[3]{h}\big)^2$ is positive for $h \ne 0$, this goes to $+\infty$ from both sides. The slopes blow up the same way on both sides, so there is a vertical tangent at $x = 0$.

For $g(x) = \sqrt[3]{x^2} = x^{2/3}$:

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

This goes to $+\infty$ as $h \to 0^+$ and to $-\infty$ as $h \to 0^-$, so there is a cusp at $x = 0$. Neither function is differentiable at $0$.

Example 3: A piecewise function that is differentiable

Is $f(x) = \begin{cases} x^2, & x \le 1 \\ 2x - 1, & x \gt 1 \end{cases}$ differentiable at $x = 1$?

Solution.

1. Continuity: the left piece gives $1^2 = 1$ and the right piece gives $2(1) - 1 = 1$. They match, and $f(1) = 1$, so $f$ is continuous at $1$.
2. Slopes: the left piece has derivative $2x$, which is $2$ at $x = 1$. The right piece has derivative $2$. They match.

So $f$ is differentiable at $x = 1$, and $f'(1) = 2$. (The line $y = 2x - 1$ is the tangent line to $y = x^2$ at $x = 1$, so the pieces join smoothly.)

Example 4: Solving for constants

Find $a$ and $b$ so that $f(x) = \begin{cases} ax^2 + 1, & x \le 2 \\ bx - 3, & x \gt 2 \end{cases}$ is differentiable at $x = 2$.

Solution. Differentiable means continuous and matching slopes, so you get two equations.

Continuity at $x = 2$: $4a + 1 = 2b - 3$.

Slopes at $x = 2$: the derivatives are $2ax$ and $b$, so $4a = b$.

Substitute $b = 4a$ into the first equation:

$$4a + 1 = 8a - 3 \quad\Rightarrow\quad 4 = 4a \quad\Rightarrow\quad a = 1, \quad b = 4$$

Check: both pieces give $5$ at $x = 2$, and both slopes are $4$.

Common mistakes

Thinking continuous means differentiable. It doesn’t. $|x|$ is continuous everywhere but has no derivative at $0$. The true statement only goes one way: differentiable implies continuous.

Checking only that the slopes match. If the pieces don’t meet, the function isn’t continuous, so it can’t be differentiable, even if the slopes of the pieces happen to agree. Always check continuity first.

Calling a vertical tangent “differentiable”. The graph does have a tangent line, but it’s vertical, so its slope (the derivative) is undefined.

Trusting the calculator at a sharp point. A graphing calculator’s numerical derivative uses a symmetric difference quotient, so it reports $0$ for the “derivative” of $|x|$ at $x = 0$. The real derivative doesn’t exist. On calculator-active AP questions, think about whether the function is differentiable before trusting the number.

Giving a name instead of a reason. “It’s a corner” earns less than “the left-hand and right-hand slopes are $-1$ and $1$, which are not equal, so $f'(2)$ does not exist.”

Practice

1. (Warm-up) True or false? Explain.

• (a) If $f$ is differentiable at $x = 3$, then $f$ is continuous at $x = 3$.
• (b) If $f$ is continuous at $x = 3$, then $f$ is differentiable at $x = 3$.
• (c) If $f$ is not continuous at $x = 3$, then $f$ is not differentiable at $x = 3$.

Solution

(a) True. Differentiability implies continuity.

(b) False. For example, $f(x) = |x - 3|$ is continuous at $3$ but has a corner there.

(c) True. This is the flipped (contrapositive) form of (a).

2. (Warm-up) Where is $f(x) = |x + 4|$ not differentiable? What does the graph look like there?

Solution

At $x = -4$. The graph is a V shape with a corner at $(-4, 0)$: the slope is $-1$ to the left and $1$ to the right.

3. (Warm-up) Explain why $f(x) = \dfrac{x^2 - 4}{x - 2}$ is not differentiable at $x = 2$.

Solution

$f(2)$ is undefined (division by zero), so $f$ is not continuous at $x = 2$. A function that isn’t continuous at a point can’t be differentiable there. (The graph is the line $y = x + 2$ with a hole at $(2, 4)$.)

4. (Core) Is $f(x) = \begin{cases} x^2 + 1, & x \lt 0 \\ x + 1, & x \ge 0 \end{cases}$ differentiable at $x = 0$? Justify.

Solution

Continuity: the left piece gives $0 + 1 = 1$ and the right piece gives $0 + 1 = 1$, and $f(0) = 1$. Continuous.

Slopes: the left piece has derivative $2x$, which is $0$ at $x = 0$. The right piece has derivative $1$. Since $0 \ne 1$, $f$ is not differentiable at $x = 0$ (there’s a corner).

5. (Core) Is $f(x) = \begin{cases} x^3, & x \le 1 \\ 3x - 2, & x \gt 1 \end{cases}$ differentiable at $x = 1$? If so, find $f'(1)$.

Solution

Continuity: $1^3 = 1$ and $3(1) - 2 = 1$, and $f(1) = 1$. Continuous.

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

So $f$ is differentiable at $x = 1$ and $f'(1) = 3$.

6. (Core) Let $f(x) = \begin{cases} x^2, & x \le 2 \\ 4x - 3, & x \gt 2 \end{cases}$. A student says: “The slopes are $2(2) = 4$ and $4$, so $f$ is differentiable at $x = 2$.” Is the student right?

Solution

No. Check continuity first: the left piece gives $4$ but the right piece gives $4(2) - 3 = 5$. The graph jumps, so $f$ is not continuous at $x = 2$, and therefore not differentiable there. Matching slopes aren’t enough.

7. (Core) Use the definition of the derivative to show that $g(x) = x|x|$ is differentiable at $x = 0$, and find $g'(0)$.

Solution

$$g'(0) = \lim_{h \to 0} \frac{h|h| - 0}{h} = \lim_{h \to 0} |h| = 0$$

The limit exists (it is $0$ from both sides), so $g$ is differentiable at $0$ and $g'(0) = 0$.

8. (Challenge) Find $a$ and $b$ so that $f(x) = \begin{cases} ax^2 + bx + 1, & x \le 1 \\ 5x - 2, & x \gt 1 \end{cases}$ is differentiable everywhere.

Solution

Each piece is a polynomial, so only $x = 1$ needs checking.

Continuity: $a + b + 1 = 5 - 2 = 3$, so $a + b = 2$.

Slopes: the derivatives are $2ax + b$ and $5$, so $2a + b = 5$.

Subtract the first equation from the second: $a = 3$. Then $b = -1$.

Check: $3 - 1 + 1 = 3$ and $5 - 2 = 3$; slopes $2(3) - 1 = 5$ and $5$.

9. (Challenge) Where is $f(x) = |x^2 - 4|$ not differentiable? Justify using one-sided slopes at one of those points.

Solution

$x^2 - 4$ changes sign at $x = \pm 2$, so those are the only candidates. Away from them, $f$ is either $x^2 - 4$ or $4 - x^2$, which are polynomials (so differentiable).

At $x = 2$: to the right, $f(x) = x^2 - 4$ with slope $2x \to 4$. To the left (between $-2$ and $2$), $f(x) = 4 - x^2$ with slope $-2x \to -4$. The one-sided slopes $4$ and $-4$ are not equal, so $f$ is not differentiable at $x = 2$. By symmetry, the same happens at $x = -2$: the slope from the left is $2(-2) = -4$ and from the right is $-2(-2) = 4$.

So $f$ is not differentiable at $x = -2$ and $x = 2$. It is continuous at both, so these are corners.