Theory

Curve sketching and optimization

How to find and classify the maxima and minima of a function with the first derivative and the second derivative, how to study concavity and inflections, how to sketch the graph with the complete curve sketching scheme and how to solve optimization problems, including those on profit, average cost and revenue.

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Maxima and minima
First derivative test

Maxima and minima

1. Absolute maximum and minimum

Let $f : A \to \mathbb{R}$ and let $x_0 \in A$ . The number $f(x_0)$ is the absolute maximum of $f$ if

f(x) \leq f(x_0) \qquad \text{for every } x \in A ,

and then $x_0$ is called an absolute maximum point; with the inequality $f(x) \geq f(x_0)$ one defines the absolute minimum and the absolute minimum point. The maximum value, if it exists, is unique, whereas the points where it is attained may be more than one. For example $f(x) = x^2$ on the interval $[-1, 2]$ has minimum $0$ , attained at $x = 0$ , and maximum $4$ , attained at $x = 2$ ; the function $\cos x$ on $\mathbb{R}$ has maximum $1$ , attained at all the points $x = 2k\pi$ with $k$ an integer.

2. Relative maximum and minimum

A relative maximum compares the value at $x_0$ only with the values at nearby points. The point $x_0 \in A$ is a relative maximum point (or local) of $f$ if there exists $\delta > 0$ such that

f(x) \leq f(x_0) \qquad \text{for every } x \in A \text{ with } |x - x_0| < \delta ,

and in the same way, with $f(x) \geq f(x_0)$ , one defines the relative minimum point. Maxima and minima are together called extrema. Every absolute maximum point is also a relative maximum point, whereas the converse is false. The function $f(x) = x^3 - 3x$ has a relative maximum at $x = -1$ , $f(-1) = 2$ , because near $-1$ the values are less than $2$ ; the maximum, however, is not absolute, because for example $f(3) = 18$ .

y = x^{3} - 3 x

The function

f (x) = x^{3} - 3 x

has a relative maximum at

x = - 1

and a relative minimum at

x = 1

; neither is absolute, because for example

f (3) = 18 > 2

Where to look for extrema

By Fermat’s theorem, at an interior point where $f$ is differentiable an extremum is possible only if the derivative vanishes. The extrema of a function defined on an interval must therefore be looked for among

the interior stationary points, where $f'(x_0) = 0$ ;
the interior points where $f$ is not differentiable;
the endpoints of the interval that belong to the domain.

All three cases actually occur. The function $|x|$ has a minimum at $0$ , where it is not differentiable, and the function $f(x) = x$ on $[0, 1]$ has a maximum at $1$ , where the derivative equals $1$ and not $0$ . The points in the list are only candidates, and the tests that follow serve to decide whether they are extrema.

First derivative test

Let $f$ be continuous on an interval $(a, b)$ containing $x_0$ and differentiable on $(a, b)$ , except possibly at the point $x_0$ .

If $f'(x) > 0$ on $(a, x_0)$ and $f'(x) < 0$ on $(x_0, b)$ , then $x_0$ is a relative maximum point.
If $f'(x) < 0$ on $(a, x_0)$ and $f'(x) > 0$ on $(x_0, b)$ , then $x_0$ is a relative minimum point.
If $f'$ has the same sign on $(a, x_0)$ and on $(x_0, b)$ , then $x_0$ is not an extremum.

In other words, a function that increases to the left of $x_0$ and decreases to the right has a maximum at $x_0$ . The test is used with the sign table of $f'$ . For $f(x) = x^3 - 3x$ one has $f'(x) = 3(x + 1)(x - 1)$ , which is positive for $x < -1$ , negative for $-1 < x < 1$ and positive for $x > 1$ ; therefore $x = -1$ is a relative maximum point and $x = 1$ a relative minimum point. The test also holds at points where $f$ is not differentiable, provided it is continuous.

Proof

The idea is to compare $f(x_0)$ with $f(x)$ through Lagrange’s theorem. Consider the first case. Let $x \in (a, x_0)$ . The function is continuous on $[x, x_0]$ and differentiable on $(x, x_0)$ , so by Lagrange’s theorem there exists $c \in (x, x_0)$ with

f(x_0) - f(x) = f'(c)\,(x_0 - x) > 0 ,

because $f'(c) > 0$ and $x_0 - x > 0$ . Now let $x \in (x_0, b)$ . In the same way there exists $c \in (x_0, x)$ with $f(x) - f(x_0) = f'(c)\,(x - x_0) < 0$ , because $f'(c) < 0$ . In both cases $f(x) < f(x_0)$ , so $x_0$ is a relative maximum point. The second case is analogous. In the third case, if for example $f' > 0$ on both sides, the same computations give $f(x) < f(x_0)$ on the left and $f(x) > f(x_0)$ on the right, hence $x_0$ is neither a maximum nor a minimum point.

Warning

A stationary point may fail to be an extremum. The function $f(x) = x^3$ has $f'(0) = 0$ , but $f'(x) = 3x^2$ is positive both to the left and to the right of $0$ ; by the first derivative test $x = 0$ is not an extremum, and indeed $f$ is increasing on all of $\mathbb{R}$ .

The theory continues with 6 more sections (concavity and inflections, second derivative test, curve sketching scheme…), plus all worked examples, the solutions and the PDF.

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