Concavity, curve sketching and optimisation – Differentiation of Polynomial, Rational and Irrational Functions and Its Applications (Senior 4 Mathematics) | Ganzaa

Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

Lessons

Senior 4 · Mathematics

Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

5 lessons

Lesson 5 of 5

Concavity, curve sketching and optimisation

25 min

The shape of a graph, and the best possible choice

The first derivative tells you where a graph rises and falls. The second derivative tells you how it bends: whether it curves up like a cup or down like a cap. With both, plus the limits and asymptotes of Unit 7, you can sketch a curve accurately without plotting a table of values. The same tools find the best value of a quantity: the largest area for a given fence, the least metal for a given volume. That is optimisation, the most widely used application of the derivative.

Definition

Concave up on an interval: the graph bends upwards (like a cup) and lies above its tangents; this happens where $f''(x) \gt 0$ , because the slopes $f^{'}$ increase.

Concave down: the graph bends downwards (like a cap) and lies below its tangents; this happens where $f''(x) \lt 0$ .

Point of inflexion: a point of the curve where the concavity changes. At such a point $f^{''} = 0$ or $f^{''}$ is undefined, and $f^{''}$ changes sign.

Example: $y = x^{3}$ : $y^{''} = 6 x$ is negative for $x \lt 0$ and positive for $x \gt 0$ , so the origin is a point of inflexion.

Important

Second derivative test. If $f^{'} (c) = 0$ and

$f''(c) \gt 0$ : $f (c)$ is a local minimum (the curve is a cup there);
$f''(c) \lt 0$ : $f (c)$ is a local maximum (a cap);
$f^{''} (c) = 0$ : the test says nothing; use the first derivative test.

Worked example 1: concavity and the second derivative test

For $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ (Lesson 4), classify the stationary points with $f^{''}$ and find the point of inflexion.

Solution $f^{'} (x) = 3 x^{2} - 6 x - 9$ , $f^{''} (x) = 6 x - 6$ . At $x = - 1$ : $f''(-1) = -12 \lt 0$ , a maximum ( $f = 10$ ). At $x = 3$ : $f''(3) = 12 \gt 0$ , a minimum ( $f = - 22$ ). $f^{''}$ changes sign at $x = 1$ : concave down for $x \lt 1$ , up for $x \gt 1$ ; the point of inflexion is $(1, f (1)) = (1, - 6)$ , halfway between the turning points.

Important

Curve sketching checklist

Domain; any symmetry (parity).
Intercepts with the axes.
Limits at the ends of the domain; vertical, horizontal and oblique asymptotes.
$f^{'}$ : stationary points and a variation table.
$f^{''}$ : concavity and points of inflexion.
Plot the key points, draw the asymptotes, and join them with the shape the tables give.

Worked example 2: a polynomial

Sketch $f (x) = x^{4} - 4 x^{3}$ .

Solution Domain $\mathbb{R}$ . Intercepts: $x^{3} (x - 4) = 0$ , so $(0, 0)$ and $(4, 0)$ . As $x \to \pm\infty$ , $f \to +\infty$ .

$f^{'} (x) = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3)$ : zero at 0 and 3. $f'(x) \lt 0$ for $x \lt 3$ (except at 0) and $\gt 0$ for $x \gt 3$ . So $x = 0$ is a stationary point with no sign change (no turning point), and $f (3) = - 27$ is the minimum.

$f^{''} (x) = 12 x^{2} - 24 x = 12 x (x - 2)$ : positive for $x \lt 0$ , negative on $(0, 2)$ , positive for $x \gt 2$ . Points of inflexion: $(0, 0)$ , a stationary inflexion, and $(2, - 16)$ .

$y = x^{4} - 4 x^{3}$ : minimum $(3, - 27)$ ; points of inflexion $(0, 0)$ (where the tangent is horizontal) and $(2, - 16)$ .

Worked example 3: a rational function

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

Solution Domain $\mathbb{R} \setminus \{1\}$ ; the only intercept is $(0, 0)$ . Dividing, $f(x) = x + 1 + \frac{1}{x - 1}$ : vertical asymptote $x = 1$ ( $f \to \pm\infty$ as $x \to 1^\pm$ ) and oblique asymptote $y = x + 1$ , with the curve above it for $x \gt 1$ and below it for $x \lt 1$ .

$f'(x) = 1 - \frac{1}{(x - 1)^2} = \frac{x(x - 2)}{(x - 1)^2}$ : local maximum $f (0) = 0$ , local minimum $f (2) = 4$ .

$f''(x) = \frac{2}{(x - 1)^3}$ : negative for $x \lt 1$ (concave down), positive for $x \gt 1$ (concave up); no inflexion, because $x = 1$ is not in the domain.

$y = \frac{x^2}{x - 1}$ : maximum $(0, 0)$ , minimum $(2, 4)$ , asymptotes $x = 1$ and $y = x + 1$ (dashed).

Worked example 4: an irrational function

Sketch $f(x) = \sqrt{x}\,(x - 3)$ , $x \ge 0$ .

Solution Intercepts $(0, 0)$ and $(3, 0)$ ; $f \to +\infty$ as $x \to +\infty$ . Write $f(x) = x^{\frac{3}{2}} - 3x^{\frac{1}{2}}$ :

f'(x) = \frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}} = \frac{3(x - 1)}{2\sqrt{x}}, \qquad f''(x) = \frac{3}{4}x^{-\frac{1}{2}} + \frac{3}{4}x^{-\frac{3}{2}} \gt 0 .

Minimum $f (1) = - 2$ ; concave up everywhere, so no inflexion. As $x \to 0^+$ , $f'(x) \to -\infty$ : the curve leaves the origin with a vertical tangent.

$y = \sqrt{x}\,(x - 3)$ : vertical tangent at the origin, minimum $(1, - 2)$ , concave up for all $x \gt 0$ .

Optimisation

To find the greatest or least value of a quantity: (1) draw a diagram and name the variables; (2) write the quantity as a function of one variable, using the constraint; (3) state the domain from the situation; (4) find the stationary points; (5) check which gives the maximum or minimum (second derivative, a variation table, or the values at the ends of the domain); (6) answer the question in words, with units.

Worked example 5: a fence against a wall

A farmer has 120 m of fencing to enclose a rectangular pen against a long straight wall (no fence is needed along the wall). Find the greatest possible area.

Solution Let the two sides perpendicular to the wall be $x$ m; the side parallel to it is $120 - 2 x$ m. Area $A (x) = x (120 - 2 x) = 120 x - 2 x^{2}$ , with $0 \lt x \lt 60$ . $A^{'} (x) = 120 - 4 x = 0$ at $x = 30$ , and $A''(x) = -4 \lt 0$ : a maximum. The pen is 30 m by 60 m, with area $1800\,\mathrm{m^2}$ .

Worked example 6: rectangle, square or circle?

(a) Show that among all rectangles with perimeter $P$ , the square has the greatest area. (b) Compare it with a circle of the same perimeter, for $P = 12\,\mathrm{m}$ .

Solution

Sides $x$ and $\frac{P}{2} - x$ : $A(x) = x\left(\frac{P}{2} - x\right)$ , $A'(x) = \frac{P}{2} - 2x = 0$ at $x = \frac{P}{4}$ , and $A'' = -2 \lt 0$ . Both sides equal $\frac{P}{4}$ : a square, area $\frac{P^2}{16}$ .
For $P = 12$ : square area $9\,\mathrm{m^2}$ . A circle of circumference 12 has radius $\frac{12}{2\pi}$ and area $\pi\left(\frac{6}{\pi}\right)^2 = \frac{36}{\pi} \approx 11.5\,\mathrm{m^2}$ , about 27% more: one reason round garden beds are a good use of a wall.

Worked example 7: the least metal for a can

A closed cylindrical can must hold $330\,\mathrm{cm^3}$ . Find the radius and height that use the least metal (surface area).

Solution $\pi r^2 h = 330$ , so $h = \frac{330}{\pi r^2}$ . Surface area $S(r) = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \frac{660}{r}$ , $r \gt 0$ .

S'(r) = 4\pi r - \frac{660}{r^2} = 0 \iff r^3 = \frac{330}{2\pi} \iff r \approx 3.74\,\mathrm{cm},

and $S''(r) = 4\pi + \frac{1320}{r^3} \gt 0$ : a minimum. Then $h = \frac{330}{\pi r^2} \approx 7.49\,\mathrm{cm}$ , exactly twice the radius (since $\pi r^3 = 165$ gives $h = \frac{330}{\pi r^2} = 2r$ ): the best can is as tall as it is wide, using about $264\,\mathrm{cm^2}$ of metal. Real drink cans are taller and narrower, so their makers are clearly weighing more than the area of metal alone.

Warning

A stationary point is not automatically the answer. Check that it is a maximum (or minimum) of the right kind, that it lies in the domain the situation allows, and compare with the values at the ends of the domain: on a closed interval the greatest value may be at an end.

Tip

In optimisation, most of the marks are for setting up: a labelled diagram, the constraint, and the quantity as a function of one variable with its domain. Once that is right, the calculus is usually a few lines.

The Organic Chemistry Tutor: Concavity, inflection points and the second derivative (about 13 min). Watch how the sign of f″ is read from a number line to give the intervals of concavity and the points of inflexion.

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Practice

Questions 1 to 4 are sketches: use the checklist. Questions 5 to 8 are optimisation problems.

Find the intervals of concavity and the point of inflexion of $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ , and classify its stationary points with $f^{''}$ .
Sketch $f (x) = x^{4} - 2 x^{2}$ , showing the turning points and points of inflexion.
Sketch $f(x) = \frac{x^2 + 4}{x}$ , showing asymptotes and turning points.
Sketch $f(x) = x\sqrt{4 - x}$ , $x \le 4$ (its turning point was found in Lesson 4).
Two positive numbers add up to 20. Find them if their product is as large as possible.
A rectangle has area $64\,\mathrm{cm^2}$ . Find the least possible perimeter.
An open box is made from a 30 cm by 20 cm card by cutting squares of side $x$ from the corners (Unit 6, Lesson 1). Find $x$ , to two decimal places, for the greatest volume.
Find the point on the curve $y = \sqrt{x}$ nearest to the point $(4, 0)$ .

Show answersHide answers

$f^{'} = 3 (x - 1) (x - 3)$ , $f^{''} = 6 x - 12$ . Max $f (1) = 5$ ( $f^{''} = - 6$ ), min $f (3) = 1$ ( $f^{''} = 6$ ). Concave down for $x \lt 2$ , up for $x \gt 2$ ; inflexion $(2, 3)$ .
Even function. $f^{'} = 4 x (x - 1) (x + 1)$ : min $(\pm 1, -1)$ , max $(0, 0)$ . $f^{''} = 12 x^{2} - 4 = 0$ at $x = \pm\frac{1}{\sqrt{3}}$ : inflexions $\left(\pm\frac{1}{\sqrt{3}}, -\frac{5}{9}\right)$ . Intercepts $0, \pm\sqrt{2}$ .
Odd function, $f(x) = x + \frac{4}{x}$ : asymptotes $x = 0$ and $y = x$ ; max $(- 2, - 4)$ , min $(2, 4)$ ; $f'' = \frac{8}{x^3}$ : concave down for $x \lt 0$ , up for $x \gt 0$ .
Intercepts $(0, 0)$ , $(4, 0)$ ; maximum $\left(\frac{8}{3}, \frac{16}{3\sqrt{3}}\right) \approx (2.67, 3.08)$ ; vertical tangent at $(4, 0)$ ; $f \to -\infty$ as $x \to -\infty$ ; concave down.
$P = x (20 - x)$ , $P^{'} = 20 - 2 x = 0$ : 10 and 10.
Sides $x$ and $\frac{64}{x}$ : $p(x) = 2x + \frac{128}{x}$ , $p' = 2 - \frac{128}{x^2} = 0$ at $x = 8$ : a square, perimeter 32 cm.
$V = 4 x^{3} - 100 x^{2} + 600 x$ , $V' = 12x^2 - 200x + 600 = 0 \Rightarrow 3x^2 - 50x + 150 = 0$ , $x = \frac{50 - \sqrt{700}}{6} \approx 3.92\,\mathrm{cm}$ (the other root, 12.74, is too big); $V \approx 1056\,\mathrm{cm^3}$ .
Distance squared $D = (x - 4)^{2} + x$ , $D^{'} = 2 x - 7 = 0$ at $x = 3.5$ : the point $(3.5, \sqrt{3.5}) \approx (3.5, 1.87)$ .

Quick check

At a stationary point x = c, f″(c) = −5. What can you conclude?

Key points

$f'' \gt 0$ : concave up; $f'' \lt 0$ : concave down; an inflexion needs $f^{''}$ to change sign.
Second derivative test at $f^{'} (c) = 0$ : $f''(c) \gt 0$ minimum, $f''(c) \lt 0$ maximum, $f^{''} (c) = 0$ no conclusion.
Sketch with the checklist: domain, symmetry, intercepts, limits and asymptotes, $f^{'}$ table, $f^{''}$ table, key points.
Optimisation: one-variable function, domain from the situation, stationary points, check the type and the ends, answer in words.

1The shape of a graph, and the best possible choice

2Optimisation

The shape of a graph, and the best possible choice

Optimisation