Curve sketching, question 28

Topic: Curve sketching
Difficulty: Capstone
Suggested time: 12 min
Marks: 6
Theory: UBC Math 100 §7 CLP-1 §3.6Key results

Let $f(x) = \dfrac{x^{2}}{x - 1}$ . Find its domain and asymptotes (vertical and slant), the intervals on which it increases and decreases, its local extrema, and the intervals on which it is concave up and concave down. Then sketch the graph.

Your answer

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

Worked solution

Strategy

Divide to reveal the linear part plus a simple remainder; every feature of the graph then follows from that remainder: where it blows up, where it flattens the slope, and which way it bends.

Working

$f(x) = x + 1 + \frac{1}{x - 1}, \qquad x \neq 1$
Division. As x → ±∞ the remainder → 0, so y = x + 1 is a slant asymptote; as x → 1⁻, f → −∞ and as x → 1⁺, f → +∞.
$f'(x) = 1 - \frac{1}{(x - 1)^{2}} = \frac{x(x - 2)}{(x - 1)^{2}}$
Zero at x = 0 and x = 2. Positive outside [0, 2], negative between (excluding 1).
$\text{local max } (0, 0),\qquad \text{local min } (2, 4)$
f′ changes + to − at 0 and − to + at 2. Note the “maximum” value 0 is below the “minimum” value 4: the asymptote sits between them.
$f''(x) = \frac{2}{(x - 1)^{3}}:\quad f'' < 0 \text{ for } x < 1,\quad f'' > 0 \text{ for } x > 1$
Concave down on the left branch, up on the right; no inflection point because concavity only changes across the asymptote.
Sketch: the left branch rises from the asymptote y = x + 1 far to the left, peaks at the origin, and dives to −∞ at x = 1; the right branch falls from +∞ at x = 1 to (2, 4) and then rises alongside y = x + 1.

Answer

\text{VA } x = 1;\quad \text{SA } y = x + 1;\quad \text{increasing on } (-\infty, 0)\cup(2, \infty);\quad \text{decreasing on } (0, 1)\cup(1, 2);\quad \max (0,0),\ \min (2,4);\quad \text{concave down for } x<1,\ \text{up for } x>1

The sketch

What earns the marks

Domain x ≠ 1; vertical asymptote x = 1 with f → −∞ from the left and +∞ from the right; slant asymptote y = x + 1 from the division f(x) = x + 1 + 1/(x − 1).
f′(x) = (x² − 2x)/(x − 1)² = x(x − 2)/(x − 1)²: increasing on (−∞, 0) and (2, ∞), decreasing on (0, 1) and (1, 2); local maximum (0, 0), local minimum (2, 4).
f″(x) = 2/(x − 1)³: concave down for x < 1, concave up for x > 1, with no inflection point (x = 1 is not in the domain).
A sketch showing both branches approaching the slant asymptote, the two extrema, and the correct behaviour at the vertical asymptote.

Check

f(−1) and f(0.5) are −0.5, below f(0) = 0, and f(1.5) and f(3) are 4.5, above f(2) = 4.

Where this goes wrong

Reporting a horizontal asymptote y = 1 from comparing leading coefficients. The numerator has higher degree, so the quotient grows without bound; dividing out gives the slant asymptote instead.