Curve sketching, question 19

Topic: Curve sketching
Difficulty: Standard
Suggested time: 4 min
Marks: 3
Theory: UBC Math 100 §7 CLP-1 §3.6Key results

Let $f(x) = \sqrt[3]{x - 2} + 1$ . Which statement describes the graph of $f$ at $x = 2$ ?

A cusp: f′(x) → −∞ as x → 2⁻ and f′(x) → +∞ as x → 2⁺, so f has a local minimum at (2, 1).
A vertical tangent line: f is continuous at 2, f′(x) → +∞ from both sides, and the concavity changes there, so (2, 1) is an inflection point.
A vertical asymptote x = 2, because f′(2) is undefined.
A horizontal tangent: x = 2 is a critical point, where f has a local extremum.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §7 CLP-1 §3.6Key results

Worked solution

Strategy

f is continuous at 2 but f′ is not defined there, so look at f′ on each side: an infinite slope of the same sign on both sides is a vertical tangent, opposite signs a cusp. Then check f″ for a change of concavity.

Working

$f(2) = \sqrt[3]{0} + 1 = 1, \qquad f'(x) = \frac{1}{3(x - 2)^{2/3}} \quad (x \neq 2)$
f is defined and continuous at 2; f′ is undefined there, so x = 2 is a singular point.
$(x - 2)^{2/3} = \bigl(\sqrt[3]{x - 2}\bigr)^{2} > 0 \ (x \neq 2) \quad\Longrightarrow\quad f'(x) \to +\infty \text{ as } x \to 2^{-} \text{ and as } x \to 2^{+}$
The slope blows up with the same sign on both sides: the tangent line is vertical, and f keeps increasing through 2, so there is no local extremum.
$f''(x) = -\frac{2}{9(x - 2)^{5/3}}:\quad f'' > 0 \text{ for } x < 2, \qquad f'' < 0 \text{ for } x > 2$
(x − 2)^5/3 has the sign of x − 2. Concave up, then concave down: (2, 1) is an inflection point.

Answer

\text{a vertical tangent at } (2, 1), \quad \text{which is also an inflection point}

Check

f(1.999) = 0.9 and f(2.001) = 1.1: f keeps rising through x = 2, with a secant slope of 100 there, steep as a vertical tangent needs.

Where this goes wrong

Reading every point where f′ is undefined as a cusp or an asymptote. Here f is continuous and f′ → +∞ from both sides, so the graph passes smoothly through (2, 1) with a vertical tangent, as y = ∛x does at the origin.

Leans on

The power rule for fractional powers, from Differentiation rules

Try a similar question

Let

f(x) = x^{2/3}(x - 5)

, defined for every real

x

. Which statement about its critical points, singular points and local extrema is correct?

My working

Saved in this browser only.

Curve sketching practice question 19