Worked examples · Sheet 3

Complete studies of rational functions

Theory used: Curve sketching scheme Even and odd functions First derivative test Sign of the second derivative and concavity Second derivative and inflections

Go to: Exercise 1 Exercise 2

Exercise 1

Study the function $f(x) = \dfrac{x^2 - 3}{x - 2}$ and sketch its graph.

Domain and symmetries.

D = \mathbb{R} \setminus \{2\}, \qquad f(-x) = \frac{x^2 - 3}{-x - 2} \neq \pm f(x)

Intersections and sign.

f(x) = 0 \iff x = \pm\sqrt{3}, \qquad f(0) = \frac{3}{2}

	$x < -\sqrt{3}$	$-\sqrt{3} < x < \sqrt{3}$	$\sqrt{3} < x < 2$	$x > 2$
$x^2 - 3$	$+$	$-$	$+$	$+$
$x - 2$	$-$	$-$	$-$	$+$
$f(x)$	$-$	$+$	$-$	$+$

Limits and asymptotes.

\lim_{x \to 2^{+}} f(x) = +\infty, \qquad \lim_{x \to 2^{-}} f(x) = -\infty

f(x) = x + 2 + \frac{1}{x - 2}, \qquad \lim_{x \to \pm\infty} \big( f(x) - (x + 2) \big) = \lim_{x \to \pm\infty} \frac{1}{x - 2} = 0

First derivative.

\begin{aligned} f'(x) &= \frac{2x\,(x - 2) - (x^2 - 3)}{(x - 2)^2} \\[6pt] &= \frac{x^2 - 4x + 3}{(x - 2)^2} = \frac{(x - 1)(x - 3)}{(x - 2)^2} \end{aligned}

	$x < 1$	$1 < x < 2$	$2 < x < 3$	$x > 3$
$f'(x)$	$+$	$-$	$-$	$+$
$f$	increasing	decreasing	decreasing	increasing

f(1) = \frac{-2}{-1} = 2, \qquad f(3) = \frac{6}{1} = 6

Second derivative.

f'(x) = 1 - \frac{1}{(x - 2)^2}, \qquad f''(x) = \frac{2}{(x - 2)^3}

The function is concave on $(-\infty, 2)$ and convex on $(2, +\infty)$ , and has no inflections. It has the vertical asymptote $x = 2$ , the oblique asymptote $y = x + 2$ , the relative maximum $(1, 2)$ and the relative minimum $(3, 6)$ .

y = \dfrac{x^{2} - 3}{x - 2}

Graph of

f (x) = \frac{x ^{2} - 3}{x - 2}

with the asymptotes

x = 2

and

y = x + 2

, the relative maximum

(1, 2)

and the relative minimum

(3, 6)

The denominator vanishes only at $x = 2$ , which is therefore the only point excluded from the domain. The domain is not symmetric with respect to $0$ , hence the function is neither even nor odd. The numerator vanishes at $\pm\sqrt{3}$ , and the sign of the quotient is read from the sign table of the two factors.

Near $2$ the numerator tends to $1$ , whereas the denominator tends to $0$ with the sign of $x - 2$ , so the line $x = 2$ is a vertical asymptote. The numerator has degree equal to the degree of the denominator plus one, hence one performs the division with remainder, which gives $x^2 - 3 = (x - 2)(x + 2) + 1$ and therefore the expression $f(x) = x + 2 + \dfrac{1}{x - 2}$ . The difference between $f$ and the line $y = x + 2$ tends to $0$ , so this line is an oblique asymptote on the right and on the left.

The derivative is computed with the quotient rule. The denominator $(x - 2)^2$ is positive on the domain, so the sign of $f'$ is that of $(x - 1)(x - 3)$ . By the first derivative test the point $x = 1$ is a relative maximum point and the point $x = 3$ a relative minimum point. The relative maximum is smaller than the relative minimum, and there is no contradiction, because the two points lie on different branches, separated by the asymptote.

For the second derivative it is convenient to differentiate the expression $f(x) = x + 2 + (x - 2)^{-1}$ , which gives $f'(x) = 1 - (x - 2)^{-2}$ and $f''(x) = 2\,(x - 2)^{-3}$ . The sign of $f''$ is that of $x - 2$ , so the function is concave on the left branch and convex on the right one. The concavity changes at $x = 2$ , which however does not belong to the domain, hence there are no inflections.

Exercise 2

Study the function $f(x) = \dfrac{x^2}{x^2 - 4}$ and sketch its graph.

Domain and symmetries.

D = \mathbb{R} \setminus \{-2, 2\}, \qquad f(-x) = \frac{(-x)^2}{(-x)^2 - 4} = f(x)

Intersections and sign.

f(x) = 0 \iff x = 0, \qquad f(x) > 0 \iff x^2 - 4 > 0 \iff |x| > 2

Limits and asymptotes.

\lim_{x \to 2^{+}} f(x) = +\infty, \qquad \lim_{x \to 2^{-}} f(x) = -\infty, \qquad \lim_{x \to \pm\infty} f(x) = 1

First derivative.

f'(x) = \frac{2x\,(x^2 - 4) - x^2 \cdot 2x}{(x^2 - 4)^2} = \frac{-8x}{(x^2 - 4)^2}

f'(x) > 0 \iff x < 0, \qquad f(0) = 0

Second derivative.

\begin{aligned} f''(x) &= \frac{-8\,(x^2 - 4)^2 + 8x \cdot 2\,(x^2 - 4) \cdot 2x}{(x^2 - 4)^4} \\[6pt] &= \frac{-8\,(x^2 - 4) + 32x^2}{(x^2 - 4)^3} = \frac{8\,(3x^2 + 4)}{(x^2 - 4)^3} \end{aligned}

f''(x) > 0 \iff x^2 - 4 > 0 \iff |x| > 2

The function is even, has the vertical asymptotes $x = \pm 2$ and the horizontal asymptote $y = 1$ ; it increases on $(-\infty, -2)$ and on $(-2, 0]$ , decreases on $[0, 2)$ and on $(2, +\infty)$ , has the relative maximum $(0, 0)$ , is convex for $|x| > 2$ , concave on $(-2, 2)$ and has no inflections.

y = \dfrac{x^{2}}{x^{2} - 4}

Graph of

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

, even, with the asymptotes

x = \pm 2

and

y = 1

and the relative maximum at the origin

The denominator $x^2 - 4$ vanishes at $\pm 2$ , which are excluded from the domain. The function contains only $x^2$ , so it is even and the graph is symmetric with respect to the $y$ -axis; it suffices to study it for $x \geq 0$ , but the computations hold anyway for every $x$ . The numerator $x^2$ is positive for $x \neq 0$ , hence the sign of $f$ is that of the denominator, positive for $|x| > 2$ .

At $x = 2$ the numerator tends to $4$ and the denominator to $0$ , positive from the right and negative from the left, so $x = 2$ is a vertical asymptote; by symmetry so is $x = -2$ , with $f \to +\infty$ as $x \to -2^{-}$ and $f \to -\infty$ as $x \to -2^{+}$ . Numerator and denominator have the same degree and the same leading coefficient, so $f(x) \to 1$ at infinity and the line $y = 1$ is a horizontal asymptote.

The denominator of the derivative is a square, positive on the domain, so $f'$ has the sign of $-8x$ . The function increases for $x < 0$ and decreases for $x > 0$ , interval by interval of the domain, and by the first derivative test $x = 0$ is a relative maximum point, with value $0$ . The maximum is not absolute, because $f$ tends to $+\infty$ near the vertical asymptotes.

In the second derivative the factor $x^2 - 4$ is canceled between numerator and denominator. The factor $3x^2 + 4$ is always positive, so the sign of $f''$ is that of $(x^2 - 4)^3$ , that is, of $x^2 - 4$ . The concavity changes at $\pm 2$ , which do not belong to the domain, hence the function has no inflections.