AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/2/2-4)
Unit 2 · Topic 2.4
2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
If a function has a derivative at a point, it must be continuous there. But a continuous function can still fail to have a derivative, at a corner, a cusp or a vertical tangent. Any discontinuity also rules out a derivative.
Key terms
- differentiable
- differentiability implies continuity
- corner
- cusp
- vertical tangent
Differentiability implies continuity
If f is differentiable at x = a, then f is continuous at x = a. Here's why: f(x) − f(a) = [(f(x) − f(a)) / (x − a)]·(x − a). As x→a, the bracket approaches f′(a) and (x − a) approaches 0, so f(x) − f(a) → f′(a)·0 = 0. That means lim (x→a) f(x) = f(a).
The contrapositive is just as useful: if f is not continuous at a, it is not differentiable at a. A jump, hole or asymptote immediately means no derivative.
Domains follow the same logic: f′ can only exist where f itself has a value. For example, ln x has no value at x = 0, so there's no slope there either, even though you might be tempted to plug 0 into a formula.
Continuity does not imply differentiability
A continuous function can still have no derivative at a point. The three standard examples:
- Corner: the left and right slopes are different finite numbers. |x| at x = 0 has slope −1 on the left and 1 on the right.
- Cusp: the graph comes to a sharp point where the slopes head to ∞ on one side and −∞ on the other. x^(2/3) at x = 0 is the example: it looks like a narrow V with curved sides.
- Vertical tangent: the tangent line is vertical, so its slope is undefined. ∛x at x = 0 is the example: the graph is continuous and smooth-looking but stands straight up at the origin.
Checking a piecewise function
For f to be differentiable at a break point c, two things must be true:
- Continuity: the pieces meet. The left-hand limit, right-hand limit and f(c) are equal.
- Matching slopes: the derivative of the left piece at c equals the derivative of the right piece at c. (This works when each piece is a differentiable formula, like a polynomial.)
| Continuous at c? | Slopes match? | Differentiable at c? |
|---|---|---|
| No | (doesn't matter) | No |
| Yes | No | No (corner) |
| Yes | Yes | Yes |
One-sided derivatives
The left-hand derivative uses h→0⁻ in the difference quotient, and the right-hand derivative uses h→0⁺. f′(c) exists only if both exist and are equal. At a corner, both exist but differ. At a cusp or vertical tangent, they are infinite.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
A corner in an absolute value
Is f(x) = |x − 2| continuous at x = 2? Is it differentiable at x = 2?
Show the solutionHide the solution
- Step 1: Continuity: f(2) = 0, and both one-sided limits are 0, so f is continuous at 2.
- Step 2: For x > 2, f(x) = x − 2 with slope 1. For x < 2, f(x) = −(x − 2) = 2 − x with slope −1.
- Step 3: The one-sided derivatives are 1 and −1. They're not equal.
Answer: f is continuous at x = 2 but not differentiable there, because the graph has a corner (left slope −1, right slope 1).
- Example 2
Find constants for differentiability
f(x) = ax² + 1 for x ≤ 1, and f(x) = bx − 2 for x > 1. Find a and b so that f is differentiable at x = 1.
Show the solutionHide the solution
- Step 1: Continuity at 1: the left piece gives a(1)² + 1 = a + 1, and the right piece approaches b − 2. Set them equal: a + 1 = b − 2.
- Step 2: Matching slopes at 1: the left piece's derivative is 2ax, which is 2a at x = 1. The right piece's derivative is b. Set 2a = b.
- Step 3: Substitute b = 2a into the first equation: a + 1 = 2a − 2, so a = 3, and b = 6.
- Step 4: Check: left piece 3x² + 1 gives 4 at x = 1, right piece 6x − 2 gives 4. Slopes: 6 and 6.
Answer: a = 3 and b = 6
- Example 3
Trap: matching slopes but not continuous
g(x) = x² for x < 1, and g(x) = 2x + 1 for x ≥ 1. A student notes that both pieces have slope 2 at x = 1 and concludes g is differentiable there. Is that right?
Show the solutionHide the solution
- Step 1: Check continuity first. The left piece approaches 1² = 1. The right piece gives g(1) = 3.
- Step 2: 1 ≠ 3, so g has a jump at x = 1 and is not continuous there.
- Step 3: Differentiability requires continuity, so the matching slopes don't matter.
Answer: No. g is not continuous at x = 1, so it can't be differentiable there.
Common mistakes
- Checking only that the slopes match at a piecewise break. Continuity must be checked first.
- Thinking “continuous” means “differentiable.” Corners, cusps and vertical tangents are continuous but not differentiable.
- Saying f is differentiable at a vertical tangent because the graph looks smooth. A vertical line has undefined slope.
On the exam
- Multiple-choice questions often show a graph and ask at which x-values f is not differentiable. Look for corners, cusps, vertical tangents and any breaks.
- Questions with constants like a and b need two equations: one from continuity and one from matching slopes.
- When a free-response question says f is differentiable, you may use that f is continuous, which unlocks IVT, EVT and MVT.
Connected topics
Videos
Check yourself
4 questions on 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist. Pick an answer to see if you got it, and why.
Which of the following functions are continuous at x = 0 but not differentiable at x = 0? I. f(x) = |x| II. g(x) = x^(1/3) III. h(x) = x|x|
Which of the following statements is true for every function f and every real number a?
Let f be the function defined by f(x) = kx² for x ≤ 2 and f(x) = mx + 4 for x > 2, where k and m are constants. If f is differentiable at x = 2, what are the values of k and m?
At which values of x is f(x) = |x² − 4| not differentiable?
0 of 4 answered