Topic 2.4 Connecting Differentiability and Continuity - Determining When Derivatives Do and Do Not Exist: explain that differentiability implies continuity but not conversely, and identify where derivatives fail to exist.

What this topic is asking

The College Board (Topic 2.4) wants you to connect differentiability and continuity: a function that is differentiable at a point must be continuous there, but a continuous function need not be differentiable. You should identify the four standard ways a derivative fails to exist - corner, cusp, vertical tangent, and discontinuity.

The one-way implication

So you can use differentiability to conclude continuity instantly, but you cannot use continuity to conclude differentiability.

Where derivatives fail to exist

Even a continuous function can lack a derivative at certain points:

• Corner: the left-hand and right-hand slopes are both finite but different (e.g. $|x|$ at $x = 0$, slopes $-1$ and $+1$).
• Cusp: the slopes approach $+\infty$ on one side and $-\infty$ on the other (e.g. $x^{2/3}$ at $x = 0$).
• Vertical tangent: the tangent line is vertical, so the slope is infinite (e.g. $x^{1/3}$ at $x = 0$).
• Discontinuity: at any jump, hole, or asymptote the function is not continuous, so it cannot be differentiable.

Testing differentiability at a point

For a piecewise function, the derivative exists at the join only if the function is continuous there and the left-hand and right-hand derivative limits agree. Both conditions are required: continuity alone is not enough (a corner is continuous but has mismatched slopes).

Recognizing non-differentiable points from a graph

On a graph, the four failure modes look distinct, and the AP exam expects you to name them on sight. A corner is a sharp point where two pieces of curve meet at different slopes, like the bottom of $|x|$. A cusp is an even sharper spike where the curve turns back on itself with the slopes diverging to $+\infty$ and $-\infty$. A vertical tangent is a smooth point where the curve momentarily runs straight up, so the slope is infinite. A discontinuity is a break, jump, or hole in the curve. At all four, no single finite tangent slope exists, so the derivative is undefined there even though three of the four (corner, cusp, vertical tangent) are continuous. Being able to scan a graph and flag exactly where $f'$ fails to exist is a common multiple-choice and free-response task.

Why this matters for the rest of calculus

Many theorems require differentiability (the Mean Value Theorem) or continuity (the Intermediate Value Theorem and Extreme Value Theorem). Knowing the one-way implication lets you check hypotheses correctly: a differentiable function automatically satisfies any continuity requirement, but you must separately verify differentiability where corners, cusps or vertical tangents might appear. For a piecewise-defined function, the standard exam task is to choose constants that make the function not just continuous but differentiable at the join, which requires two equations: the pieces must meet (continuity) and their slopes must match (equal one-sided derivatives). Solving both conditions together is a direct application of the idea that differentiability is strictly stronger than continuity.