The function f has a relative maximum at c if there is some interval (r, s) (even a very small one) containing c for which f(c) ≥ f(x) for all x between r and s for which f(x) is defined.

f has a relative minimum at c if there is an interval (r, s) (even a very small one) containing c for which f(c) ≤ f(x) for all x between r and s for which f(x) is defined.

By a relative exremum, we mean either a relative maximum or a relative minimum.

The following figure shows several relative extrema.

Note Our definition of relative extremum allows f to have a relative extremum at an endpoint of its domain; the definitions used in some textbooks do not.

Let

f(x) = x² - 2x,   with domain [0, 4].

Here is its graph.

Looking at the graph, we find that f has:

• A relative maximum at (0, 0);
• A relative minimum at (1, - 1);
• A relative maximum at (4, 8).

Relative extrema may sometimes also be absolute extrema, as the following definition shows.

f has an absolute maximum at c if f(c) ≥ f(x) for every x in the domain of f.

f has an absolute minimum at c if f(c) ≤ f(x) for every x in the domain of f.

The following figure shows several relative extrema that are also absolute extrema.

Note All absolute extrema are automatically relative extrema, according to our convention.

Once again, let

f(x) = x² - 2x,   with domain [0, 4].

Looking again at its relative extrema, we see that:

• The maximum at (0, 0) is not an absolute maximum;
• The minimum at (1, -1) is an absolute minimum;
• The maximum at (4, 8) is an absolute maximum.

Note If we changed the domain to [0, +∞), then there would be no absolute maximum. (Why?)

If f is continuous on its domain and differentiable except at a few isolated points, then its relative extrema occur among the following types of points.

1. Stationary points: points xin the domain where f'(x) = 0. To locate stationary points, set f'(x) = 0 and solve for x.
2. Singular points: points x in the domain where f'(x) is not defined. To locate singular points, find values x where f'(x) is not defined, but f(x) is defined.
3. Endpoints: the endpoints of the domain, if any. Recall that closed intervals contain endpoints, but open intervals do not.

The following figure shows several instances of all three types.



Still uncomfortable with this material? Try the on-line tutorial on maxima and minima.

1. Let us look yet again at the graph of f(x) = x² - 2x,   with domain [0, 4].

 
More Examples
Stationary points: Let f(x) = x³ - 12x.
Then to locate the stationary points, set f'(x) = 0 and solve for x. This gives 3x²- 12 = 0, so x = ±2 are the stationary points.

Singular points: Let f(x) = 3(x- 1)^1/3.
Then f'(x) = (x- 1)^{- 2/3} = 1/(x- 1)^2/3.
f'(x) is not defined when x = 1, although f(x) is defined when x = 1. Thus, the (only) singular point is x = 1.

Endpoints: Let f(x) = 1/x, with domain (- ∞, 0) [1, +∞).
Then the only endpoint in the domain of f is x = 1. On the other hand, the natural domain of 1/x has no endpoints.

Note If we changed the domain to [0, +∞), then there would be no absolute maximum. (Why?)

We solve optimization problems of the following form: Find the values of the unknowns x, y, . . . maximizing (or minimizing) the value of the objective function f, subject to certain constraints. The constraints are equations and inequalities relating or restricting the variables x, y, . . . .

To solve such a problem, we use the constraint equations to write all of the variables in terms of one chosen variable, substitute these into the objective function f, and then find extrema as above. (We use any constraint inequalities to determine the domain of the resulting function of one variable.) Specifically:

1. Identify the unknown(s).
These are usually the quantities asked for in the problem.

2. Identify the objective function.
This is the quantity you are asked to maximize or minimize.

3. Identify the constraint(s).
These can be equations relating variables or inequalities expressing limitations on the values of variables.

4. State the optimization problem.
This will have the form "Maximize [minimize] the objective function subject to the constraint(s)."

5. Eliminate extra variables.
If the objective function depends on several variables, solve the constraint equations to express all variables in terms of one particular variable. Substitute these expressions into the objective function to rewrite it as a function of a single variable. Substitute the expressions into any inequality constraints to help determine the domain of the objective function.

6. Find the absolute maximum (or minimum) of the objective function.
Use the techniques described above.

Here is a maximization problem:

Maximize A = xy	Objective Function
subject to x + 2y = 100, x ≥ 0, and y ≥ 0	Constraints

Let us carry out the procedure for solving. Since we already have the problem stated as an optimization problem, we can start at Step 5.

5. Eliminate extra variables.
We can do this by solving the constraint equation x + 2y = 100 for x (getting x = 100 - 2y) and substituting in the objective function and the inequality involving x:

A = xy = (100- 2y)y = 100y - 2y²
(100- 2y) ≥ 0,   or   y ≤ 50.

6. Find the absolute maximum (or minimum) of the objective function.
Following the above procedure, we get two endpoints and a stationary point with values as shown.

From the table, we see that the largest possible value of A is 1,250, which occurs when y = 25. The corresponding value of x is x = 100 - 2y, so x = 50 when y = 25.

Acceleration
The acceleration of a moving object is the derivative of its velocity, i.e., the second derivative (derivative of the derivative) of its position function.

Concavity
A curve is concave up if its slope is increasing, in which case the second derivative will be positive. A curve is concave down if its slope is decreasing, in which case the second derivative will be negative. A point where the graph of f changes concavity, from concave up to concave down or vice versa, is called a point of inflection. At a point of inflection the second derivative will either be undefined or 0.

Acceleration
If t is time in hours and the position of a car at time t is s(t) = t³ + 2t² miles, then:

Velocity = v(t) = s'(t) = 3t² + 4t miles per hour.
Acceleration = a(t) = s" (t) = 6t + 4 miles per hour per hour.

Concavity
Consider f(x) = x³ - 3x, whose graph is shown below.

f"(x) = 6x is negative when x < 0 and positive when x > 0. The graph of f is concave down when x < 0 and concave up when x > 0. f has a point of inflection at x = 0, where the second derivative is 0.

We can use graphing technology to draw a graph, but we need to use calculus to understand what we are seeing. The most interesting features of a graph are the following.

Features of a Graph
1. The x- and y-intercepts If y = f(x), find the x-intercept(s) by setting y = 0 and solving for x; find the y-intercept by setting x = 0.

2. Relative extrema Use the techniques described above to locate the relative extrema.

3. Points of inflection Set f"(x) = 0 and solve for x to find candidate points of inflection.

4. Behavior near points where the function is not defined If f(x) is not defined at x = , consider lim_{x → -} f(x) and lim_{x → +} f(x) to see how the graph of f approaches this point.

5. Behavior at infinity Consider lim_{x → -∞} f(x) and lim_{x → +∞} f(x) if appropriate, to see how the graph of f behaves far to the left and right.

If you have Excel, try our Excel First and Second Derivative Graphing Utility to see plots of any function and its first two derivatives.

Here is the graph of

f(x)

=

x2 (x+1)(x- 2)

To analyze this, we follow the procedure at left:

1. The x- and y-intercepts Setting y = 0 and solving for x gives x = 0. This is the only x-intercept. Setting x = 0 and solving for y gives y = 0: the y-intercept.

2. Relative extrema The only extrema are stationary points found by setting f'(x) = 0 and solving for x, giving x = 0 and x = - 4. The corresponding points on the graph are the relative maximum (0, 0) and the relative minimum at (- 4, 8/9).

3. Points of inflection Solving f"(x) = 0 analytically is difficult, so we can solve it numerically (plot the second derivative and estimate where is crosses the x-axis) and find that the point of inflection lies at x - 6.1072.

4. Behavior near points where the function is not defined The function is not defined at x = - 1 and x = 2. The limits as x approaches these values from the left and right can be inferred from the graph:

x	lim →	- 1-	f(x)	=	+∞
x	lim →	- 1+	f(x)	=	- ∞
x	lim →	2-	f(x)	=	- ∞
x	lim →	2+	f(x)	=	+∞

5. Behavior at infinity From the graph (or the function) we see that

x	lim →	- ∞	f(x)	=	1
x	lim →	+∞	f(x)	=	1.

If Q is a quantity that is varying with time, then the rate at which Q is changing is given by the time derivative, dQ/dt. A typical related rates problem asks for the rate of change of a quantity Q, given the rates of changes of various other related quantities.

Solving a Related Rates Problem

A. The Problem

1. List the related, changing quantities.
2. Restate the problem in terms of rates of change. Rewrite the problem using mathematical notation for the changing quantities and their derivatives

B. The Relationship

1. Draw a diagram, if appropriate, showing the changing quantities.
2. Find an equation or equations relating the changing quantities.
3. Take the derivative with respect to time of the equation(s) relating the quantities to get the derived equation(s), relating the rates of change of the quantities.

C. The Solution

Substitute into the derived equation(s) the given values of the quantities and their derivatives, then solve for the derivative required.

Take a look at the on-line tutorial on related rates.

The traffic at the RealWorld web site is given by

h = - 0.001p² + 400     hits per day,

The elasticity of demand, E, is the percentage rate of decrease of demand per percentage increase in price. We obtain it from the demand equation according to the following formula:

E

=

-

dq dp

.

p q

.

where the demand equation expresses demand, q, as a function of unit price, p. We say that the demand is elastic if E > 1, the demand is inelastic if E < 1, and the demand has unit elasticity if E = 1.

To find the unit price that maximizes revenue, we express E as a function of p, set E = 1, and then solve for p.

Suppose that the demand equation is q = 20,000 - 2p. Then

E

=

- (- 2)

p 20,000- 2p

=

p 10,000- p

If p = 2,000, then E = 1/4, and demand is inelastic at this price.

If p = 8,000, then E = 4, and demand is elastic at this price.

If p = 5,000, then E = 1, and the demand has unit elasticity at this price.