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Differentiability & Continuity – Problem of the Week

Differentiability and Continuity – Calculus Problem of the Week

This Calculus Problem of the Week is a typical exam question that tests your understanding of differentiability versus continuity, and your ability to recognize the Definition of the Derivative and what it tells you. It’s also typical of a “which of these statements is/are true?” question. See how you do!

Practice Problem: Differentiability versus Continuity


Students encountering a question such as this one for the first time on an exam are sometimes intimidated. (So it’s great that you’re seeing it now!)

Here’s an important tip: Focus on one statement at a time and decide whether it is true or false based on the information provided. That last bit is important, because if we cannot definitively conclude that a statement is true based on what we are told, then the statement is automatically false.

Let examine each Statement one-by-one.

Examining: “I. f is differentiable at x = 1.”

Recall the Definition of the Derivative at $x = a$:

Definition of the Derivative

\[f'(a) = \lim_{h \to 0}\frac{f(a+h) – f(a)}{h} \]

The condition we are given for the function, \[\overbrace{\lim_{h \to 0}\frac{f(1+h) -f(1) }{h}}^{f'(1)} = 0 \] is thus actually the derivative of f at $x=1,$ $f'(1),$ which you simply have to recognize (through practice). Furthermore, the statement tells us that $f'(1) = 0.$

Since we are told that $f'(1) = 0,$ f is certainly differentiable at x = 1, and so Statement I is true:

✔ I. f is differentiable at x = 1.

Examining: “II. f is continuous as x = 1.”

For a function to be differentiable at a given x-value, it must be continuous there. That is, “differentiability implies continuity.” Since we concluded above that f is differentiable at x = 1, it must therefore also be continuous there. Statement II is thus true:

✔ II. f is continuous as x = 1.

Examining: “III. f is continuous at x = 0.”

The information provided simply tells us nothing about the function’s behavior at x = 0: the function could be continuous there, or might not be, and could even be undefined at x = 0! Since these various possibilities exist and we can’t reach a definite conclusion, the statement is automatically (in the worlds of propositional logic and mathematics) false:

✗ III. f is continuous at x = 0.

Summarizing, we have

✔ I. f is differentiable at x = 1.
✔ II. f is continuous as x = 1.
✗ III. f is continuous at x = 0.

and so the answer is (c) I and II only $\; \cmark$.

What thoughts or questions does this problem leave you with? Or what other questions about differentiability and continuity (or other Calculus topics) do you have? Please let us know below, or on our Forum!

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