Worked examples · Sheet 1

Studies of rational functions

Theory used: Scheme of curve sketching Monotonicity test First derivative test Convexity and second derivative Sign chart of f' and f''Qualitative graph

Go to: Exercise 1 Exercise 2

Exercise 1

Study the function

f (x) = \frac{x ^{2}}{x - 1}

and sketch its qualitative graph.

Domain. The denominator must be nonzero, therefore $D = (- \infty, 1) \cup (1, + \infty)$ .

Symmetries and periodicity. $f (- x) = \frac{x ^{2}}{- x - 1}$ is different from both $f (x)$ and $- f (x)$ , therefore the function is neither even nor odd; it is not periodic.

Sign and intersections. The numerator $x^{2}$ is positive for $x \neq = 0$ and zero at $0$ , therefore the sign of $f$ coincides with the sign of $x - 1$ . Hence $f (x) > 0$ for $x > 1$ , $f (x) < 0$ for $x < 1$ with $x \neq = 0$ , and the graph meets the axes only at the origin.

Limits and asymptotes. Polynomial division gives

f (x) = x + 1 + \frac{1}{x - 1} .

At the endpoints of the domain:

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

therefore $x = 1$ is a vertical asymptote. Since $f (x) - (x + 1) = \frac{1}{x - 1} \to 0$ as $x \to \pm \infty$ , the line $y = x + 1$ is an oblique asymptote on both sides.

Continuity. $f$ is a quotient of polynomials with nonzero denominator on $D$ , therefore it is continuous on $D$ .

Differentiability and monotonicity. $f$ is differentiable on all of $D$ and

f^{'} (x) = 1 - \frac{1}{( x - 1 ) ^{2}} = \frac{( x - 1 ) ^{2} - 1}{( x - 1 ) ^{2}} = \frac{x ( x - 2 )}{( x - 1 ) ^{2}} .

The denominator is positive on $D$ , therefore $f^{'} (x) > 0$ for $x < 0$ or $x > 2$ . By the monotonicity test $f$ is increasing on $(- \infty, 0]$ , decreasing on $[0, 1)$ , decreasing on $(1, 2]$ , increasing on $[2, + \infty)$ .

Maxima and minima. The stationary points are $0$ and $2$ . At $0$ the derivative changes from positive to negative, therefore by the first derivative test $0$ is a relative maximum point with $f (0) = 0$ ; at $2$ the derivative changes from negative to positive, therefore $2$ is a relative minimum point with $f (2) = 4$ . Since $f$ tends to $+ \infty$ and to $- \infty$ , the absolute maximum and minimum do not exist.

Concavity and inflection points. Differentiating $f^{'} (x) = 1 - (x - 1)^{- 2}$ gives

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

negative for $x < 1$ and positive for $x > 1$ . The function is concave on $(- \infty, 1)$ and convex on $(1, + \infty)$ ; the change of concavity occurs at $x = 1$ , which does not belong to the domain, therefore there are no inflection points.

Graph.

y = \dfrac{x^{2}}{x - 1}

y = x + 1

y = \frac{x ^{2}}{x - 1}

: maximum at

(0, 0)

, minimum at

(2, 4)

, vertical asymptote

x = 1

and oblique asymptote

y = x + 1

(dashed)

The division with remainder of the numerator by the denominator, $x^{2} = (x - 1) (x + 1) + 1$ , is the step that simplifies the whole study. Since the difference $f (x) - (x + 1)$ is $\frac{1}{x - 1}$ and tends to $0$ , the line $y = x + 1$ is an oblique asymptote without the need to compute separately the limits of $\frac{f ( x )}{x}$ and of $f (x) - m x$ . The same expression makes the second derivative immediate.

For the one-sided limits at $1$ the numerator tends to $1 > 0$ , while the denominator tends to $0$ changing sign: from the left $x - 1 < 0$ , therefore the quotient tends to $- \infty$ ; from the right $x - 1 > 0$ , therefore it tends to $+ \infty$ . The result is consistent with the sign already found, since $f < 0$ to the left of $1$ and $f > 0$ to the right.

The value of the relative maximum, $f (0) = 0$ , is less than the value of the relative minimum, $f (2) = 4$ . No rule requires the opposite: the two points lie on branches separated by the vertical asymptote, and on each branch the function reaches arbitrarily large and arbitrarily small values.

Exercise 2

Study the function

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

and sketch its qualitative graph.

Domain. The denominator vanishes at $0$ , therefore $D = (- \infty, 0) \cup (0, + \infty)$ .

Symmetries and periodicity. $f (- x) = \frac{x ^{2} + 4}{- x} = - f (x)$ , therefore the function is odd and the graph is symmetric with respect to the origin; it is not periodic.

Sign and intersections. The numerator $x^{2} + 4$ is always positive, therefore the sign of $f$ is the sign of $x$ . We have $f (x) > 0$ for $x > 0$ and $f (x) < 0$ for $x < 0$ ; the function never vanishes and the graph does not meet the axes.

Limits and asymptotes. It is convenient to write

f (x) = x + \frac{4}{x} .

At the endpoints of the domain:

x \to \pm \infty lim f (x) = \pm \infty, x \to 0^{-} lim f (x) = - \infty, x \to 0^{+} lim f (x) = + \infty,

therefore $x = 0$ is a vertical asymptote. Since $f (x) - x = \frac{4}{x} \to 0$ as $x \to \pm \infty$ , the line $y = x$ is an oblique asymptote on both sides.

Continuity. $f$ is a quotient of polynomials with nonzero denominator on $D$ , therefore it is continuous on $D$ .

Differentiability and monotonicity. $f$ is differentiable on $D$ and

f^{'} (x) = 1 - \frac{4}{x ^{2}} = \frac{x ^{2} - 4}{x ^{2}} .

Therefore $f^{'} (x) > 0$ for $∣ x ∣ > 2$ . By the monotonicity test $f$ is increasing on $(- \infty, - 2]$ , decreasing on $[- 2, 0)$ , decreasing on $(0, 2]$ , increasing on $[2, + \infty)$ .

Maxima and minima. The stationary points are $- 2$ and $2$ . By the first derivative test $- 2$ is a relative maximum point with $f (- 2) = - 4$ , while $2$ is a relative minimum point with $f (2) = 4$ , in agreement with the odd symmetry. The absolute maximum and minimum do not exist, because the function is unbounded in both directions.

Concavity and inflection points. Differentiating $f^{'} (x) = 1 - 4 x^{- 2}$ gives

f^{''} (x) = \frac{8}{x ^{3}},

negative for $x < 0$ and positive for $x > 0$ . The function is concave on $(- \infty, 0)$ and convex on $(0, + \infty)$ ; the point $0$ does not belong to the domain, therefore there are no inflection points.

Graph.

y = x + \dfrac{4}{x}

y = x

y = \frac{x ^{2} + 4}{x}

: maximum at

(- 2, - 4)

, minimum at

(2, 4)

, vertical asymptote

x = 0

and oblique asymptote

y = x

(dashed)

The odd symmetry, verified in the second step, halves the work: every result found for $x > 0$ carries over to $x < 0$ by changing the sign of both coordinates. The relative minimum $(2, 4)$ gives the relative maximum $(- 2, - 4)$ , and convexity on $(0, + \infty)$ gives concavity on $(- \infty, 0)$ .

On the right branch the function is positive, decreases down to the value $4$ and then increases toward $+ \infty$ along the asymptote $y = x$ : the value $4$ is the minimum of that branch, that is $f (x) \geq 4$ for every $x > 0$ . It is not, however, an absolute minimum, because on the left branch the function takes negative values of arbitrarily large absolute value.