Examining a function for extremal values: second method

From (8.1), it is clear that in the vicinity of a maximum value of $f(x)$, in passing along the graph from left to right, $f'(x)$ changes from $+$ to 0 to $-$. Hence $f'(x)$ is a decreasing function, and by §8.3 we know that its derivative, i.e. the second derivative ($= f''(x)$) of the function itself, is negative or zero.

Similarly, we have, from (8.2), that in the vicinity of a minimum value of $f(x)$ $f'(x)$ changes from $-$ to 0 to $+$. Hence $f'(x)$ is an increasing function and by §8.3 it follows that $f''(x)$ is positive or zero.

The student should observe that $f''(x)$ is positive not only at minimum values but also at ``nearby'' points, $P$ say, to the right of such a critical point. For, as a point passes through P in moving from left to right, slope $= \tan \tau = \frac{dy}{dx} = f'(x)$ is an increasing function. At such a point the curve is said to be concave upwards. Similarly, $f''(x)$ is negative not only at maximum points but also at ``nearby ''points, $Q$ say, to the left of such a critical point. For, as a point passes through $Q$, slope $= \tan \tau = \frac{dy}{dx} = f'(x)$ is a decreasing function. At such a point the curve is said to be concave downwards.

At a point where the curve is concave upwards we sometimes say that the curve has a ``positive bending,]] and where it is concave downwards a ``negative bending.''

We may then state the sufficient conditions for maximum and minimum values of $f(x)$ for certain values of the variable as follows:

$$f(x)\ {\rm is\ a\ maximum\ if\ }f'(x) = 0\ {\rm and}\ f''(x) =\ {\rm a\ negative\ number}.$$ (8.3)

$$f(x)\ {\rm is\ a\ minimum\ if}\ f'(x) = 0\ {\rm and}\ f''(x) =\ {\rm a\ positive\ number}.$$ (8.4)

Following is the corresponding working rule.

• FIRST STEP. Find the first derivative of the function.

• SECOND STEP. Set the first derivative equal to zero and solve the resulting equation for real roots in order to find the critical values of the variable.

• THIRD STEP. Find the second derivative.

• FOURTH STEP. Substitute each critical value for the variable in the second derivative. If the result is negative, then the function is a maximum for that critical value; if the result is positive, the function is a minimum.

When $f''(x) = 0$, or does not exist, the above process fails, although there may even then be a maximum or a minimum; in that case the first method given in the last section still holds, being fundamental. Usually this second method does apply, and when the process of finding the second derivative is not too long or tedious, it is generally the shortest method.

Example 8.6.1   Let us now apply the above rule to test analytically the function

$$M = x^2 + \frac{432}{x}$$

found in Example 8.1.2.

Solution. $f(x) =\ x^2 + \frac{432}{x}$.

First step. $f'(x) =\ 2x - \frac{432}{x^2}$.

Second step. $2x - \frac{432}{x^2} =\ 0$.

Third step. $f''(x) =\ 2 + \frac{864}{x^3}$.

Fourth step. $f''(6) =\ +$. Hence $f(6) = 108$, minimum value.

In SAGE:

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]


sage: x = var("x")
sage: f(x) = x^2 + 432/x
sage: f1(x) = diff(f(x),x); f1(x)
2*x - 432/x^2
sage: f2(x) = diff(f(x),x,2); f2(x)
864/x^3 + 2
sage: crit_pts = solve(f1(x) == 0,x); crit_pts
[x == 3*sqrt(3)*I - 3, x == -3*sqrt(3)*I - 3, x == 6]
sage: x0 = crit_pts[2].rhs(); x0
6
sage: f2(x0)
6
sage: f(x0)
108

This tells us that $x_0=6$ is a critical point and that $f''(x_0)>0$, so it is a minimum.

The work of finding maximum and minimum values may frequently be simplified by the aid of the following principles, which follow at once from our discussion of the subject.

(a)

The maximum and minimum values of a continuous function must occur alternately,

(b)

When $c$ is a positive constant, $c \cdot f(x)$ is a maximum or a minimum for such values of $x$, and such only, as make $f(x)$ a maximum or a minimum.

Hence, in determining the critical values of x and testing for maxima and minima, any constant factor may be omitted.

When $c$ is negative, $c \cdot f(x)$ is a maximum when $f(x)$ is a minimum, and conversely.

(c)

If $c$ is a constant, $f(x)$ and $c + f(x)$ have maximum and minimum values for the same values of $x$.

Hence a constant term may be omitted when finding critical values of $x$ and testing.

In general we must first construct, from the conditions given in the problem, the function whose maximum and minimum values are required, as was done in the two examples worked out in §8.1. This is sometimes a problem of considerable difficulty. No rule applicable in all cases can be given for constructing the function, but in a large number of problems we may be guided by the following

General directions.

(a)

Express the function whose maximum or minimum is involved in the problem.

(b)

If the resulting expression contains more than only variable, the conditions of the problem will furnish enough relations between the variables so that all may be expressed in terms of a single one.

(c)

To the resulting function of a single variable apply one of our two rules for finding maximum and minimum values.

(d)

In practical problems it is usually easy to tell which critical value will give a maximum and which a minimum value, so it is not always necessary to apply the fourth step of our rules.

(e)

Draw the graph of the function in order to check the work.