Curve sketching, question 16

Topic: Curve sketching
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §7 CLP-1 §3.6Key results

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?

Its only critical point is x = 2, a local minimum, and f has no other local extremum.
x = 2 is a critical point, a local minimum, and x = 0 is a vertical asymptote, because f′ is undefined there.
x = 0 is a singular point but not a local extremum, because f′ is negative on both sides of it; x = 2 is a critical point, a local minimum.
x = 0 is a singular point, where f has a local maximum (a cusp), and x = 2 is a critical point, where f has a local minimum.

How sure are you?

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

Worked solution

Strategy

Find where f′ is zero and where it is undefined, then let one sign chart classify both kinds of point.

Working

$f(x) = x^{5/3} - 5x^{2/3} \quad\Longrightarrow\quad f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5(x - 2)}{3\sqrt[3]{x}}$
Power rule, then put both terms over the common denominator 3∛x.
$f'(2) = 0; \qquad f'(0) \text{ undefined, while } f(0) = 0$
x = 2 is a critical point; x = 0 is a singular point, since f is defined there but its derivative is not.
$x < 0:\ \frac{(-)}{(-)} > 0; \qquad 0 < x < 2:\ \frac{(-)}{(+)} < 0; \qquad x > 2:\ \frac{(+)}{(+)} > 0$
Sign chart of x − 2 over ∛x; the cube root has the sign of x.
$\text{local max } f(0) = 0, \qquad \text{local min } f(2) = -3\sqrt[3]{4}$
Rising then falling at 0, with f′ → +∞ from the left and −∞ from the right: a cusp. Falling then rising at 2.

Answer

x = 0:\ \text{singular, local max (cusp)}; \qquad x = 2:\ \text{critical, local min}

Check

f(−0.1) ≈ −1.10 and f(0.1) ≈ −1.06 are below f(0) = 0, and f(2) ≈ −4.76 is below f(1) = −4 and f(3) ≈ −4.16.

Where this goes wrong

Solving f′(x) = 0 and stopping at x = 2. A point where f is defined but f′ is not, here x = 0, can also be a local extremum, so it belongs in the sign chart.