The definition of the derivative, question 22

Topic: The definition of the derivative
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §3.3 CLP-1 §2.2Key results

Let $f$ be the function below. Show that $f$ is continuous at $x = 1$ . Then use one-sided difference quotients, from the definition of the derivative, to decide whether $f$ is differentiable at $x = 1$ .

f(x) = \begin{cases} x^{2} + 1, & x \leq 1 \\[4pt] \dfrac{4}{x + 1}, & x > 1 \end{cases}

Write your reasoning in full, then compare it against the model solution.

Your answer

Theory UBC Math 100 §3.3 CLP-1 §2.2Key results

Worked solution

Strategy

Differentiability at a join needs continuity first, then equal one-sided limits of the difference quotient. Each piece supplies f(1 + h) on its own side, and f(1) always comes from the piece that contains x = 1.

Working

$f(1) = 1^{2} + 1 = 2, \qquad \lim_{x \to 1^{+}} \frac{4}{x + 1} = \frac{4}{2} = 2$
The left piece gives f(1) and the left-hand limit; the right piece tends to the same value, so f is continuous at 1.
$\lim_{h \to 0^{-}} \frac{(1 + h)^{2} + 1 - 2}{h} = \lim_{h \to 0^{-}} \frac{2h + h^{2}}{h} = \lim_{h \to 0^{-}} (2 + h) = 2$
The slope from the left, from the definition.
$\frac{\frac{4}{2 + h} - 2}{h} = \frac{4 - 2(2 + h)}{h(2 + h)} = \frac{-2h}{h(2 + h)} = \frac{-2}{2 + h}$
For h > 0 the right piece applies; combine the fractions and cancel h.
$\lim_{h \to 0^{+}} \frac{-2}{2 + h} = -1$
The slope from the right.
The one-sided limits are 2 and −1. They differ, so the limit that defines f′(1) does not exist: the graph is continuous at 1 but has a corner there.

Answer

\text{continuous at } 1; \quad \text{one-sided slopes } 2 \text{ and } -1, \quad \text{so } f'(1) \text{ does not exist}

What earns the marks

Checks continuity: f(1) = 2, and 4/(x + 1) tends to 4/2 = 2 as x → 1⁺, so both one-sided limits equal f(1).
Computes the left-hand quotient from the definition: ((1 + h)² + 1 − 2)/h = 2 + h, which tends to 2 as h → 0⁻.
Computes the right-hand quotient from the definition, with f(1) = 2: (4/(2 + h) − 2)/h = −2/(2 + h), which tends to −1 as h → 0⁺.
Concludes that the one-sided limits 2 and −1 differ, so f′(1) does not exist: f is continuous but not differentiable at 1.

Check

Chords from (1, 2) have slope 1.99 to x = 0.99 and about −0.995 to x = 1.01, heading for 2 and −1, not one common value.

Where this goes wrong

Skipping the continuity check, or taking f(1 + h) from the wrong piece. For h > 0 the quotient uses 4/(2 + h), while f(1) = 2 still comes from x² + 1, the piece that holds at x = 1.

Leans on

One-sided limits, from Limits

Try a similar question

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The definition of the derivative practice question 22