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B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Curve Sketching III: Rational Functions

B1.3 B1.5About 75 minutes
Learning goals
  • I can sketch a simple rational function using its domain, intercepts, symmetry and asymptotes together with f′ and f″.
  • I can include vertical asymptotes in sign charts and write intervals correctly on each side of them.
  • I can verify a rational-function sketch with graphing technology.

Words to know

branch
One connected piece of a graph between vertical asymptotes.
oblique asymptote
A slanted line y = mx + b approached by the graph as x → ±∞.

The full toolkit

In MHF4U you sketched rational functions using intercepts, asymptotes and a few test points. You could not be sure where the turning points were, or how the branches bent. With f′ and f″ you can.

The routine is the polynomial routine plus asymptotes:

  1. Domain (exclude zeros of the denominator), holes and intercepts.
  2. Symmetry (even or odd).
  3. Asymptotes: vertical (with one-sided behaviour), horizontal or oblique (with above/below).
  4. f′: critical numbers, increase/decrease, local extrema.
  5. f″: concavity and inflection points.
  6. Sketch, then verify with technology.

Put every x-value not in the domain into your sign charts. The sign of f′ or f″ can change there even though there is no point on the graph.

Worked example

Example 1: An oblique asymptote and two turning points

Sketch f(x) = x²/(x − 1).

  1. Domain: x ≠ 1. Intercepts: f(0) = 0; f = 0 only at x = 0 (a double zero, so the graph touches the axis).
  2. Asymptotes: VA x = 1 (numerator 1 ≠ 0). Division: x² = (x − 1)(x + 1) + 1, so f(x) = x + 1 + 1/(x − 1). OA y = x + 1; the graph is above it for x > 1 and below it for x < 1.
  3. Near x = 1: lim (x→1⁺) f = ∞ and lim (x→1⁻) f = −∞ (from 1/(x − 1)).
  4. f′(x) = 1 − 1/(x − 1)² = [(x − 1)² − 1]/(x − 1)² = x(x − 2)/(x − 1)². Critical numbers 0 and 2. Signs: +, −, −, + (borders 0, 1, 2). Max (0, 0); min (2, 4), since f(2) = 4/1.
  5. f″(x) = 2/(x − 1)³. Negative for x < 1 (concave down), positive for x > 1 (concave up). No inflection point.
  6. Second derivative test check: f″(0) = −2 < 0 (max ✓), f″(2) = 2 > 0 (min ✓).

See the chart and graph below: max (0, 0), min (2, 4), VA x = 1, OA y = x + 1, concave down left of 1, concave up right of 1.

f(x) = x²/(x − 1)x012f′(x)+−−+0undef.0f″(x)−−++undef.f(x)maxVAmin
The vertical asymptote x = 1 goes into the chart like a critical number, but it is not a point of the graph. The curved arrows show each piece: rising ∩, falling ∩, falling ∪, rising ∪.
y = x²/(x − 1)xy−4−3−2−1123456−8−426x = 1y = x + 1maxmin (2, 4)
The complete sketch: local max (0, 0), local min (2, 4), vertical asymptote x = 1 and oblique asymptote y = x + 1 (dashed). The left branch is concave down and the right branch is concave up.
Worked example

Example 2: Two vertical asymptotes

Sketch f(x) = 1/(x² − 4).

  1. Domain: x ≠ ±2. Intercepts: y-int −1/4; no x-intercepts (the numerator is never 0).
  2. Symmetry: f(−x) = f(x): even.
  3. Asymptotes: VAs x = ±2. HA y = 0. For |x| > 2, f > 0 (above the axis); for |x| < 2, f < 0.
  4. Near x = 2: right side f → ∞, left side f → −∞. By symmetry, near x = −2: left side f → ∞, right side f → −∞.
  5. f′(x) = −2x(x² − 4)⁻² = −2x/(x² − 4)². Critical number x = 0. f′ > 0 for x < 0 and f′ < 0 for x > 0 (x ≠ ±2). Local max (0, −1/4).
  6. f″: product rule on −2x(x² − 4)⁻² gives −2(x² − 4)⁻² + 8x²(x² − 4)⁻³ = [−2(x² − 4) + 8x²]/(x² − 4)³ = (6x² + 8)/(x² − 4)³.
  7. The numerator is always positive, so f″ has the sign of (x² − 4)³: positive for |x| > 2 (concave up), negative for |x| < 2 (concave down). No inflection points.

VAs x = ±2, HA y = 0, local max (0, −1/4); concave up on (−∞, −2) and (2, ∞), concave down on (−2, 2).

y = 1/(x² − 4)xy−4−3−2−11234−3−2−112x = −2x = 2max (0, −1/4)
An even function with two vertical asymptotes and the horizontal asymptote y = 0 (the x-axis). The middle branch is concave down with a local maximum (0, −1/4); the outer branches are concave up.
Worked example

Example 3: No vertical asymptote, three inflection points

Sketch f(x) = 2x/(x² + 1).

  1. Domain: all real numbers (x² + 1 > 0). Intercepts: only (0, 0).
  2. Symmetry: f(−x) = −f(x): odd.
  3. Asymptotes: no VA; HA y = 0 (degree 1 < degree 2). f > 0 for x > 0, f < 0 for x < 0.
  4. f′(x) = [2(x² + 1) − 2x(2x)]/(x² + 1)² = 2(1 − x²)/(x² + 1)². Critical numbers ±1; signs −, +, −. Min (−1, −1), max (1, 1).
  5. f″(x) = 4x(x² − 3)/(x² + 1)³ (worked out in the challenge of the lesson on finding the second derivative). Zeros 0, ±√3; signs −, +, −, + (borders −√3, 0, √3).
  6. f(±√3) = ±2√3/4 = ±√3/2 ≈ ±0.87.

Min (−1, −1), max (1, 1); IPs (−√3, −√3/2), (0, 0), (√3, √3/2); HA y = 0. Concave down on (−∞, −√3) and (0, √3); concave up on (−√3, 0) and (√3, ∞).

y = 2x/(x² + 1)xy−5−4−3−2−112345−1.5−0.51max (1, 1)min (−1, −1)IPIPIP
An odd function with no vertical asymptote. It falls from the HA y = 0 to its minimum (−1, −1), rises through the origin to its maximum (1, 1), then falls back toward y = 0, and has three inflection points: (−√3, −√3/2), (0, 0) and (√3, √3/2).

Rational-function traps

  • Forgetting to put vertical asymptotes in the f′ and f″ charts. The sign often changes there.
  • Writing one interval across an asymptote, such as "decreasing on (0, 2)" for Example 1. Write (0, 1) and (1, 2).
  • Calling a change of concavity at an asymptote an inflection point.
  • Expanding the quotient-rule numerator but not factoring it.
Just for fun

A rational function was asked why it never shows up at x = 1. It replied, "I have a strict policy of asymptotic distancing."

Hands-on activity

Predict, sketch, verify

You need: graph paper, a pencil, and a free graphing tool (Desmos or GeoGebra) or a graphing calculator.

  1. Without technology, sketch y = (x² + 3)/(x − 1). Find the VA, the oblique asymptote, both turning points and the concavity.
  2. Write your predictions in a small table.
  3. Graph the function with technology. Zoom out to see the asymptotes, then zoom in near the turning points.
  4. Mark each prediction as correct or fix it.

What to look for (answers): x² + 3 = (x − 1)(x + 1) + 4, so f = x + 1 + 4/(x − 1): VA x = 1, OA y = x + 1. f′ = 1 − 4/(x − 1)² = (x − 3)(x + 1)/(x − 1)²: max (−1, −2), min (3, 6). f″ = 8/(x − 1)³: concave down for x < 1, up for x > 1.

Your turn

1

For f(x) = x²/(x + 2), find the asymptotes.

2

For f(x) = x²/(x + 2), find and classify the critical points.

3

For f(x) = x²/(x + 2), describe the concavity.

4

For f(x) = 1/(x² + 1), give the symmetry, asymptote and extremum.

5

For f(x) = x/(x − 2), find the intervals of increase/decrease.

6

For f(x) = 4/(x² − 1), find the local extremum and its type.

More practice: warm-up

1

Domain of f(x) = 3/(x² − 9)?

2

Is 1/(x² − 4) even or odd?

3

Is 2x/(x² + 1) even or odd?

4

Should a vertical asymptote appear in the sign chart of f′?

5

Can an inflection point occur at a vertical asymptote?

More practice: on your own

1

Sketch f(x) = (x² + 4)/x: asymptotes, extrema, concavity.

2

Sketch f(x) = x/(x² − 1): symmetry, asymptotes, monotonicity.

3

For f(x) = x/(x² − 1), find the inflection point. (f″ = 2x(x² + 3)/(x² − 1)³.)

4

Sketch f(x) = (x − 1)/(x + 1)²: asymptotes and extremum.

5

Find the intervals of concavity of f(x) = (x² + 4)/x.

Challenge

1

For f(x) = (x − 1)/(x + 1)², find f″ and the inflection point.

2

Sketch f(x) = x³/(x² − 1) by finding its asymptotes and critical points.

3

A cylindrical can has volume 500 cm³. Its surface area as a function of radius r is A(r) = 2πr² + 1000/r. Sketch A for r > 0 and find the radius with the least surface area. (This is a preview of optimization in Applications of Derivatives.)

Check your understanding

Quiz

1.

For f(x) = x²/(x − 1), what is the local minimum?

2.

Where is f(x) = 1/(x² − 4) concave up?

3.

Which statement about f(x) = 2x/(x² + 1) is true?

4.

For f(x) = x + 1/x, which is the correct description of where f decreases?

5.

The concavity of a rational function changes from down to up across its vertical asymptote x = 3. At x = 3 there is:

6.

What is the oblique asymptote of f(x) = (x² + 3)/(x − 1)?

For parents and teachers

Big idea (B1.3, B1.5): sketching a rational function combines everything in the unit: asymptotes and one-sided limits, f′ for turning points and f″ for concavity. Vertical asymptotes must go into every sign chart.

How to help: the predict-sketch-verify activity is ideal. Ask your teen to show you the predictions table before they open the graphing tool.

Common difficulties:

  • Algebra with the quotient rule (simplify and factor before reading signs).
  • Joining intervals across an asymptote.
  • Mixing up the behaviour on the two sides of a vertical asymptote.
  • Forgetting the oblique asymptote when the numerator's degree is one higher.