Curve sketching, question 28: UBC MATH 100 practice

Question 28 of 30Challenging4 marks · 8 minTheory UBC Math 100 §7.6CLP-1 §3.6Key results

Sketch the graph of the function below. Find its domain, intercepts and symmetry, its asymptotes, the intervals on which it increases and decreases with its local extrema, and the intervals on which it is concave up and concave down with its inflection points.

f(x) = \frac{x^{2} - 4}{x^{2} + 4}

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

Your answer

Theory UBC Math 100 §7.6 CLP-1 §3.6Key results

Worked solution

Strategy

Gather what the formula says directly, then let the sign of f′ give the rise and fall and the sign of f″ the bending. Evenness halves the work: what happens for x > 0 is mirrored for x < 0.

Working

$x^{2} + 4 > 0 \implies \text{domain } \mathbb{R}, \qquad f(-x) = f(x)$
No vertical asymptotes, and the graph is symmetric about the y-axis.
$f(0) = -1, \qquad f(x) = 0 \iff x = \pm 2$
The intercepts.
$f(x) = 1 - \frac{8}{x^{2} + 4} \to 1 \text{ as } x \to \pm\infty, \qquad f(x) < 1$
Horizontal asymptote y = 1 at both ends, approached from below.
$f'(x) = \frac{2x\left(x^{2} + 4\right) - 2x\left(x^{2} - 4\right)}{\left(x^{2} + 4\right)^{2}} = \frac{16x}{\left(x^{2} + 4\right)^{2}}$
The quotient rule. f′ has the sign of x, so f decreases on (−∞, 0) and increases on (0, ∞): f(0) = −1 is the local, and absolute, minimum.
$f''(x) = \frac{16\left(4 - 3x^{2}\right)}{\left(x^{2} + 4\right)^{3}}$
The quotient rule again, after cancelling a factor x² + 4. f″ > 0 for |x| < 2/√3 and f″ < 0 for |x| > 2/√3.
$f\!\left(\pm\frac{2}{\sqrt{3}}\right) = \frac{\frac{4}{3} - 4}{\frac{4}{3} + 4} = -\frac{1}{2}$
The concavity changes at x = ±2/√3, so these are the inflection points.
Sketch: from just below y = 1 far to the left, the graph falls, concave down, through (−2, 0) to the inflection point (−2/√3, −1/2), then concave up to the minimum (0, −1); the right half is its mirror image.

Answer

\text{decreasing on } (-\infty, 0),\ \text{increasing on } (0, \infty);\quad \min (0, -1);\quad \text{HA } y = 1;\quad \text{concave up on } \left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right),\ \text{down outside};\quad \text{inflection points } \left(\pm\frac{2}{\sqrt{3}}, -\frac{1}{2}\right)

The sketch

What earns the marks

Domain all real x, since x² + 4 > 0, so no vertical asymptote; f is even, so the graph is symmetric about the y-axis; intercepts (0, −1) and (±2, 0); horizontal asymptote y = 1 at both ends, approached from below.
f′(x) = 16x/(x² + 4)²: f decreases on (−∞, 0) and increases on (0, ∞), with a local and absolute minimum f(0) = −1 and no local maximum.
f″(x) = 16(4 − 3x²)/(x² + 4)³: concave up on (−2/√3, 2/√3), concave down outside it, with inflection points (±2/√3, −1/2).
A sketch consistent with all of this: a trough at (0, −1) that rises through (±2, 0) and flattens towards y = 1 at both ends.

Check

f(1) = −3/5 and f(3) = 5/13, so the graph rises from −1 through 0 at x = 2 towards 1; f″(1) = 16/125 > 0 and f″(2) = −1/4 < 0, either side of 2/√3 ≈ 1.155.

Where this goes wrong

Stopping at the first derivative. The bend of the graph, concave up near the minimum and concave down as it flattens towards y = 1, comes only from the second derivative, and the inflection points are where its sign changes.

The idea behind it

The curve sketching lesson, a figure to move, in a few steps

Leans on

The quotient rule, from Differentiation rules

Try a similar question

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.

My working

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Curve sketching practice question 28