Lesson 10: Differentiability & Continuity

(a) Opening Question

Consider the absolute-value function f(x) = |x|. Its graph is a sharp "V" with the corner sitting at the origin.

Sketch it, then ask: what is the slope at x = 0?

Walk in from the left, along y = −x. The slope there is −1 the whole way. Now walk in from the right, along y = x. The slope there is +1 the whole way. The two sides demand two different slopes at the single point x = 0.

There is no single tangent line that fits both — the graph turns a hard corner. So f(x) = |x| has no derivative at x = 0, even though the graph never lifts off the page: it is perfectly continuous there.

That is the heart of this lesson. Continuity is about the graph having no breaks. Differentiability is a stronger demand: the graph must also be smooth — no corners, no cusps, no vertical tangents. This lesson pins down exactly when a derivative fails to exist, and how to say so in AP language.

(b) Core Concepts

A function can be continuous (no gaps, jumps, or holes) yet still fail to be differentiable (no well-defined tangent slope). Understanding the precise relationship between these two ideas — and the four ways differentiability breaks — is a recurring AP exam theme.

One-sided derivatives

The derivative at a point is a two-sided limit:

f'(c) = lim_{h→0} [f(c + h) − f(c)] / h

Like any two-sided limit, it exists only if the left-hand and right-hand pieces agree. Define the one-sided derivatives:

left-hand derivative:   f'₋(c) = lim_{h→0⁻} [f(c + h) − f(c)] / h
right-hand derivative:  f'₊(c) = lim_{h→0⁺} [f(c + h) − f(c)] / h

f is differentiable at c if and only if f'₋(c) and f'₊(c) both exist (as finite numbers) and are equal. This single test diagnoses every case below. For f(x) = |x|, f'₋(0) = −1 and f'₊(0) = +1; they exist but disagree, so f'(0) does not exist.

The theorem: differentiable ⇒ continuous

Theorem. If f is differentiable at x = c, then f is continuous at x = c.

In words: differentiability is the stronger condition. Smoothness implies connectedness. Intuitively, if the difference quotient [f(c+h) − f(c)]/h is to approach a finite slope as h → 0, the numerator f(c+h) − f(c) must shrink to 0 — which is exactly the statement that lim_{x→c} f(x) = f(c), i.e. continuity.

The converse is FALSE

The theorem does not run backwards. Continuity does not imply differentiability. The honest way to state this is through the contrapositive, which IS always true and is your most useful exam tool:

Contrapositive. If f is not continuous at x = c, then f is not differentiable at x = c.

So a discontinuity is an automatic disqualification: no tangent line where the graph breaks. But the reverse trap is the one AP loves — a function can be continuous everywhere and still fail to be differentiable somewhere. The counterexample is f(x) = |x| at x = 0: continuous (no break), but not differentiable (a corner). Memorize this counterexample — it is the standard proof that "continuous ⇒ differentiable" is false.

The four ways differentiability fails

A function fails to be differentiable at x = c in exactly one of four ways:

1. Corner — the one-sided derivatives both exist but are unequal (f'₋ ≠ f'₊). The graph abruptly changes slope. Example: |x| at 0.
2. Cusp — the one-sided slopes head to infinity in opposite directions (one to +∞, the other to −∞). The graph comes to a sharp point. Example: f(x) = x^(2/3) at 0, where f'(x) = (2/3)x^(−1/3) → −∞ from the left and → +∞ from the right.
3. Vertical tangent — the slope tends to ±∞ from both sides (the same infinite sign), so the tangent line is vertical and has no real slope. Example: f(x) = x^(1/3) at 0, where f'(x) = (1/3)x^(−2/3) → +∞ from both sides.
4. Discontinuity — any hole, jump, or vertical asymptote. By the contrapositive, no derivative exists where f is not continuous.

Note that cases 1–3 are all continuous points (the graph is unbroken) where differentiability still fails — they are the witnesses that the converse is false.

• Panel 1 (CORNER): f(x) = |x| near x = 0. A sharp "V": the line y = −x for x < 0 meeting the line y = x for x ≥ 0 at the origin (0,0). Continuous (the two pieces touch), but the left slope is −1 and the right slope is +1 — slopes do not match. Label "corner: f'₋ = −1 ≠ f'₊ = +1, not differentiable, but continuous."
• Panel 2 (CUSP): f(x) = x^(2/3) near x = 0. A continuous curve coming down to a sharp point at the origin (0,0) and back up, like a narrow valley pinched to a spike. Tangent direction is nearly vertical on both sides but the curve approaches from below-left and below-right. Label "cusp: slopes → −∞ (left) and +∞ (right), not differentiable, but continuous."
• Panel 3 (VERTICAL TANGENT): f(x) = x^(1/3) near x = 0. A continuous increasing S-shaped curve passing through the origin (0,0) with a vertical tangent line there; the curve is steep and nearly vertical at the origin. Label "vertical tangent: slope → +∞ from both sides, not differentiable, but continuous."
• Panel 4 (DISCONTINUITY): a function with a jump at x = c. Solid segment ending at an open circle at (c, 1) on the left piece, and a filled dot at (c, 3) starting the right piece. The graph breaks. Label "jump discontinuity: not continuous ⇒ not differentiable (contrapositive)."]

Making a piecewise function differentiable — the classic "find a and b" problem

A favorite AP problem hands you a piecewise function with unknown constants and asks you to choose them so the function is differentiable at the seam. The seam must satisfy two independent conditions:

1. Continuity — the two pieces must meet: their values agree at the seam (left value = right value).
2. Matching derivatives — the two pieces must meet smoothly: their one-sided derivatives agree at the seam (left slope = right slope).

Two conditions, two unknowns — solve the system. Skipping condition 2 is the single most common error on this problem type.

Worked example. Find a and b so that

        ⎧ x²        x ≤ 1
f(x) =  ⎨
        ⎩ ax + b    x > 1

is differentiable at x = 1.

Condition 1 — continuity at x = 1. The pieces must agree at x = 1:

left value:  (1)² = 1
right value: a(1) + b = a + b
⇒  a + b = 1

Condition 2 — matching derivatives at x = 1. Differentiate each piece, then equate the one-sided slopes at x = 1:

left piece  f(x) = x²    ⇒ f'(x) = 2x   ⇒ f'₋(1) = 2(1) = 2
right piece f(x) = ax+b  ⇒ f'(x) = a    ⇒ f'₊(1) = a
⇒  a = 2

Solve the system. From condition 2, a = 2. Substitute into condition 1: 2 + b = 1, so b = −1.

a = 2,   b = −1

So f(x) = x² for x ≤ 1 and f(x) = 2x − 1 for x > 1 is differentiable at the seam. Check: continuity, 2(1) − 1 = 1 = 1² ✓; slopes, both equal 2 ✓. The line 2x − 1 is exactly the tangent line to x² at x = 1, which is the geometric meaning of "joining smoothly."

Calculator notes

A graphing calculator is excellent for spotting trouble: graph the function and look for corners, cusps, and breaks. But nDeriv is dangerous near non-differentiable points. nDeriv computes a symmetric difference quotient [f(c+h) − f(c−h)]/(2h) with a small built-in h, so it will happily return a number even where no derivative exists:

TI-84: nDeriv(abs(X), X, 0) → 0   (but f'(0) does NOT exist for |x|!)

For |x|, the symmetric quotient averages the −1 and +1 slopes to a fake 0. nDeriv at a corner, cusp, or jump returns a meaningless value. Trust the graph and the one-sided derivatives, not the nDeriv number, near any suspected non-differentiable point.

(c) Worked Examples

Example 1 — Classify non-differentiable points from a graph [NO CALC]

Problem. A continuous function g is graphed on [−4, 4]. It has a sharp corner at x = −2, a vertical tangent at x = 1, and a sharp downward spike (cusp) at x = 3. Elsewhere it is a smooth curve. At which x-values is g not differentiable, and why?

Strategy. Scan for corners, cusps, vertical tangents, and discontinuities. Name the reason for each.

Solution. g is not differentiable at x = −2, x = 1, and x = 3.

Justification.

• At x = −2, g has a corner: the left- and right-hand derivatives both exist but are unequal, so g'(−2) does not exist.
• At x = 1, g has a vertical tangent: the slope tends to ±∞, so g'(1) does not exist (the tangent line is vertical and has no finite slope).
• At x = 3, g has a cusp: the one-sided slopes tend to infinity in opposite directions, so g'(3) does not exist.
• g is differentiable everywhere else on [−4, 4], since the graph is smooth (and continuous) there.

Example 2 — A corner via one-sided derivatives [NO CALC]

Problem. Let f(x) = |x − 3|. Show that f is continuous at x = 3 but not differentiable there.

Strategy. Continuity: check the limit equals the value. Differentiability: compute both one-sided derivatives and compare.

Solution. Rewrite as a piecewise function:

f(x) = |x − 3| = ⎧ −(x − 3) = 3 − x,   x < 3
                 ⎨
                 ⎩  (x − 3) = x − 3,   x ≥ 3

Continuity at x = 3: the left limit is 3 − 3 = 0, the right limit is 3 − 3 = 0, and f(3) = 0. All three agree, so f is continuous at x = 3.

One-sided derivatives at x = 3:

left piece  3 − x   ⇒ f'₋(3) = −1
right piece x − 3   ⇒ f'₊(3) = +1

Justification. f is continuous at x = 3 because lim_{x→3} f(x) = 0 = f(3). However, f is not differentiable at x = 3 because the left-hand derivative f'₋(3) = −1 and the right-hand derivative f'₊(3) = +1 are not equal. This is a corner — and a direct demonstration that continuity does not imply differentiability.

Example 3 — Solve a piecewise differentiability problem [NO CALC]

Problem. Find constants c and d so that

        ⎧ cx + d        x < 2
g(x) =  ⎨
        ⎩ x² − 1        x ≥ 2

is differentiable at x = 2.

Strategy. Impose both conditions at the seam x = 2: equal values (continuity) and equal one-sided slopes (matching derivatives). Solve the resulting system.

Solution.

Matching derivatives at x = 2: f'₋(2) = c (slope of cx + d); f'₊(2) = 2x |_{x=2} = 4 (since the derivative of x² − 1 is 2x). Set equal:

c = 4

Continuity at x = 2: left value c(2) + d = 2c + d; right value 2² − 1 = 3. Set equal:

2c + d = 3

Substitute c = 4: 2(4) + d = 3 ⇒ 8 + d = 3 ⇒ d = −5.

c = 4,   d = −5

Justification. With c = 4 and d = −5, g is continuous at x = 2 because both pieces equal 3 there, and g is differentiable at x = 2 because both one-sided derivatives equal 4. Both conditions are required: continuity alone (any c, d with 2c + d = 3) would only guarantee the graph connects, not that it is smooth.

Example 4 — Justify differentiability and continuity [NO CALC]

Problem. A function h is continuous on [0, 5] and has a jump discontinuity removed — actually h is continuous everywhere on [0, 5] except it has a corner at x = 2 and a removable hole patched at x = 4 (so h is continuous at x = 4). State, with justification, whether h is differentiable at x = 2 and at x = 4.

Strategy. Apply the definitions and the contrapositive directly.

Solution & Justification.

• At x = 2: h is not differentiable, because h has a corner there — the left- and right-hand derivatives are unequal. (Continuity at x = 2 is not enough; smoothness fails.)
• At x = 4: the hole has been patched so h is continuous there, but continuity alone does not guarantee differentiability. If the curve is smooth at x = 4 (one-sided slopes match), then h is differentiable there; we cannot conclude differentiability from continuity alone. This is the precise point: differentiable ⇒ continuous, but continuous ⇏ differentiable.

(d) Common Mistakes

Assuming continuous ⇒ differentiable. Students see an unbroken graph and conclude the derivative exists everywhere. Why it's wrong: the implication only runs one way — differentiable ⇒ continuous, never the reverse. Fix: a continuous graph can still have corners (|x|), cusps (x^(2/3)), or vertical tangents (x^(1/3)). Always check for smoothness, not just connectedness.

On a piecewise problem, checking only continuity and forgetting derivative-matching. Students set the two pieces equal at the seam, solve, and stop — getting a function that connects but bends. Why it's wrong: differentiability requires the one-sided derivatives to match too. Fix: always impose both conditions (equal values AND equal slopes). One equation per condition; two equations for two unknowns.

Misreading a vertical tangent as differentiable. A vertical tangent looks like the graph has a perfectly good (steep) tangent line, so students say the derivative exists. Why it's wrong: a vertical line has no real slope — the derivative is ±∞, which means it does not exist as a finite number. Fix: "vertical tangent" is a reason the derivative fails to exist, not a reason it succeeds.

Trusting nDeriv at a corner or jump. Students compute nDeriv(abs(X), X, 0), get 0, and report f'(0) = 0. Why it's wrong: nDeriv uses a symmetric difference and averages the mismatched one-sided slopes into a meaningless number, even where no derivative exists. Fix: near any suspected non-differentiable point, use one-sided derivatives by hand; never trust the nDeriv value.

Confusing "the limit exists" with "the derivative exists." Students check that lim_{x→c} f(x) exists (continuity) and call the function differentiable. Fix: differentiability is about the limit of the difference quotient existing, which is strictly more than the function's limit existing.

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## (e) Practice Problems

Which statement is always true?

- (A) If f is continuous at x = c, then f is differentiable at x = c.

- (B) If f is differentiable at x = c, then f is continuous at x = c.

- (C) If f is not continuous at x = c, then f may still be differentiable at x = c.

- (D) Continuity and differentiability are equivalent.

The function f(x) = |x + 4| fails to be differentiable at:

- (A) x = 4 (B) x = −4 (C) x = 0 (D) nowhere

A function has a vertical tangent at x = c. This means at x = c the function is:

- (A) discontinuous and not differentiable

- (B) continuous and differentiable

- (C) continuous but not differentiable

- (D) differentiable but not continuous

Which feature does NOT make a continuous function non-differentiable at a point?

- (A) a corner (B) a cusp (C) a vertical tangent (D) a horizontal tangent

At a corner, the left- and right-hand derivatives:

- (A) are both infinite (B) both fail to exist

- (C) exist but are unequal (D) exist and are equal

For f(x) = x^(1/3), the derivative f'(x) = (1/3)x^(−2/3) as x → 0 behaves so that f has at x = 0 a:

- (A) corner (B) cusp (C) vertical tangent (D) jump discontinuity

If g(x) = x² for x ≤ 0 and g(x) = mx for x > 0 is differentiable at x = 0, then m =

- (A) 2 (B) 1 (C) 0 (D) −1

Consider the graph below. At which x-value is the function continuous but not differentiable?

• At x = −1: a smooth rounded local minimum (differentiable)
• At x = 2: a sharp corner where the graph changes direction abruptly (continuous, not differentiable)
• At x = 4: an open circle at (4, 3) and a filled dot at (4, 1) — a jump discontinuity (not continuous)
• Elsewhere the curve is smooth]
• (A) x = −1 (B) x = 2 (C) x = 4 (D) the function is differentiable everywhere

A student computes nDeriv(abs(X−2), X, 2) and the calculator returns 0. The correct conclusion is:

- (A) f'(2) = 0

- (B) f has a horizontal tangent at x = 2

- (C) f'(2) does not exist; nDeriv is unreliable at the corner

- (D) f is differentiable at x = 2 with slope 0

If f is differentiable at x = 5, which MUST be true?

- (A) f is continuous at x = 5

- (B) f has a horizontal tangent at x = 5

- (C) f(5) = 0

- (D) f'(5) = 0

(Short answer.) Find constants a and b so that

f(x) = ⎧ ax + b     x < 1
       ⎨
       ⎩ x³         x ≥ 1

is differentiable at x = 1. Show both conditions.

(Short answer.) Let f(x) = |2x − 6|. Find the one-sided derivatives at x = 3 and state whether f is differentiable there.

(Justification.) A function f is continuous at x = 0 but lim_{h→0⁻}[f(0+h)−f(0)]/h = 3 while lim_{h→0⁺}[f(0+h)−f(0)]/h = −1. Is f differentiable at x = 0? Justify your answer using precise language.

(Justification.) A student writes: "Since g(x) = x^(2/3) is continuous at x = 0, it is differentiable at x = 0." Identify the error and write a correct justification.

The piecewise function f(x) = x² for x ≤ 1 and f(x) = 2x + k for x > 1 is continuous at x = 1. The value of k is:

- (A) 1 (B) −1 (C) 0 (D) 2

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## (f) AP Exam Focus — Free Response Question

This is the lesson's full AP-style FRQ. Total: 9 points. No calculator. All justification must be shown in writing using precise AP language.

Problem. Let f be the function defined by

        ⎧ x² + 1        x ≤ 2
f(x) =  ⎨
        ⎩ ax + b        x > 2

where a and b are constants.

(a) (2 points) Find the value of b in terms of a so that f is continuous at x = 2. Show the reasoning.

(b) (3 points) Find the values of a and b so that f is differentiable at x = 2. Show that your function satisfies both the continuity and the derivative conditions.

(c) (2 points) Let g be the function g(x) = |x − 2|. State, with justification, whether g is differentiable at x = 2.

(d) (2 points) A student claims, "Every continuous function is differentiable." Use part (c) to explain, with correct mathematical language, why this claim is false.

Model Solution

(a) For f to be continuous at x = 2, the two pieces must agree at x = 2:

left value:  (2)² + 1 = 5
right value: a(2) + b = 2a + b

Setting them equal: 2a + b = 5, so

b = 5 − 2a

(b) Differentiability at x = 2 requires both continuity (from part (a)) and matching one-sided derivatives.

Derivative condition. The left piece x² + 1 has derivative 2x, so f'₋(2) = 2(2) = 4. The right piece ax + b has derivative a, so f'₊(2) = a. Setting these equal:

a = 4

Continuity condition. Substitute a = 4 into b = 5 − 2a from part (a):

b = 5 − 2(4) = 5 − 8 = −3

So a = 4 and b = −3.

Verification.

• Continuity: left value 2² + 1 = 5; right value 4(2) + (−3) = 8 − 3 = 5. They agree ✓.
• Derivatives: f'₋(2) = 4 and f'₊(2) = 4. They agree ✓.

Therefore f is differentiable at x = 2 with a = 4, b = −3.

(c) Write g(x) = |x − 2| as a piecewise function:

g(x) = ⎧ 2 − x,   x < 2   ⇒ g'₋(2) = −1
       ⎨
       ⎩ x − 2,   x ≥ 2   ⇒ g'₊(2) = +1

g is not differentiable at x = 2, because the left-hand derivative g'₋(2) = −1 and the right-hand derivative g'₊(2) = +1 are not equal. (Note g is continuous at x = 2 since both pieces equal 0 there — but a corner prevents differentiability.)

(d) The claim is false. The function g(x) = |x − 2| from part (c) is continuous at x = 2 (its left and right limits both equal g(2) = 0), yet it is not differentiable at x = 2 because its one-sided derivatives −1 and +1 are unequal — a corner. The correct relationship is one-directional: differentiability implies continuity, but continuity does not imply differentiability. The function g is a counterexample to the student's claim.

Scoring Commentary — where students lose points

• Part (a) — 2 points: 1 point for setting the two pieces equal at x = 2 (the continuity condition 2a + b = 5), 1 point for the correct answer b = 5 − 2a. Students lose a point for plugging in x = 2 to only one piece, or for using a strict-inequality value incorrectly.
• Part (b) — 3 points: 1 point for the derivative condition a = 4 (correctly differentiating both pieces and equating one-sided slopes), 1 point for b = −3, 1 point for showing both conditions are satisfied. The classic loss: solving only a = 4 without finding b, or finding a and b but never demonstrating the continuity check. Students who match derivatives but forget continuity (or vice versa) cap at 2 of 3.
• Part (c) — 2 points: 1 point for computing the two one-sided derivatives −1 and +1, 1 point for the conclusion "not differentiable because the left- and right-hand derivatives are unequal." Students lose points for saying only "it has a corner" without naming the unequal one-sided derivatives, or for claiming the derivative is 0 (a common nDeriv-style error).
• Part (d) — 2 points: 1 point for identifying g as continuous yet not differentiable at x = 2, 1 point for stating the correct one-way implication (differentiable ⇒ continuous, but not conversely). Students lose points for a vague answer like "it has a sharp point" without the words continuous and not differentiable, or for reversing the implication.

a = 3,   b = −2

f(x) = ⎧ 6 − 2x,  x < 3  ⇒ f'₋(3) = −2
       ⎨
       ⎩ 2x − 6,  x ≥ 3  ⇒ f'₊(3) = +2