## 122. Maxima and Minima.

We shall say that the value $$\phi(\xi)$$ assumed by $$\phi(x)$$ when $$x = \xi$$ is a maximum if $$\phi(\xi)$$ is greater than any other value assumed by $$\phi(x)$$ in the immediate neighbourhood of $$x = \xi$$, if we can find an interval $${[\xi – \delta, \xi + \delta]}$$ of values of $$x$$ such that $$\phi(\xi) > \phi(x)$$ when $$\xi – \delta < x < \xi$$ and when $$\xi < x < \xi + \delta$$; and we define a minimum in a similar manner. Thus in the figure the points $$A$$ correspond to maxima, the points $$B$$ to minima of the function whose graph is there shown. It is to be observed that the fact that $$A_{3}$$ corresponds to a maximum and $$B_{1}$$ to a minimum is in no way inconsistent with the fact that the value of the function is greater at $$B_{1}$$ than at $$A_{3}$$.

Theorem C. A necessary condition for a maximum or minimum value of $$\phi(x)$$ at $$x = \xi$$ is that $$\phi'(\xi) = 0$$.1

This follows at once from Theorem A. That the condition is not sufficient is evident from a glance at the point $$C$$ in the figure. Thus if $$y = x^{3}$$ then $$\phi'(x) = 3x^{2}$$, which vanishes when $$x = 0$$. But $$x = 0$$ does not give either a maximum or a minimum of $$x^{3}$$, as is obvious from the form of the graph of $$x^{3}$$ (Fig. 10, § 23).

But there will certainly be a maximum at $$x = \xi$$ if $$\phi'(\xi) = 0$$, $$\phi'(x) > 0$$ for all values of $$x$$ less than but near to $$\xi$$, and $$\phi'(x) < 0$$ for all values of $$x$$ greater than but near to $$\xi$$: and if the signs of these two inequalities are reversed there will certainly be a minimum. For then we can (by Cor. 3 of § 121) determine an interval $${[\xi – \delta, \xi]}$$ throughout which $$\phi(x)$$ increases with $$x$$, and an interval $${[\xi, \xi + \delta]}$$ throughout which it decreases as $$x$$ increases: and obviously this ensures that $$\phi(\xi)$$ shall be a maximum.

This result may also be stated thus. If the sign of $$\phi'(x)$$ changes at $$x = \xi$$ from positive to negative, then $$x = \xi$$ gives a maximum of $$\phi(x)$$: and if the sign of $$\phi'(x)$$ changes in the opposite sense, then $$x = \xi$$ gives a minimum.

## 123.

There is another way of stating the conditions for a maximum or minimum which is often useful. Let us assume that $$\phi(x)$$ has a second derivative $$\phi”(x)$$: this of course does not follow from the existence of $$\phi'(x)$$, any more than the existence of $$\phi'(x)$$ follows from that of $$\phi(x)$$. But in such cases as we are likely to meet with at present the condition is generally satisfied.

Theorem D. If $$\phi'(\xi) = 0$$ and $$\phi”(\xi) \neq 0$$, then $$\phi(x)$$ has a maximum or minimum at $$x = \xi$$, a maximum if $$\phi”(\xi) < 0$$, a minimum if $$\phi”(\xi) > 0$$.

Suppose, , that $$\phi”(\xi) < 0$$. Then, by Theorem A, $$\phi'(x)$$ is negative when $$x$$ is less than $$\xi$$ but sufficiently near to $$\xi$$, and positive when $$x$$ is greater than $$\xi$$ but sufficiently near to $$\xi$$. Thus $$x = \xi$$ gives a maximum.

## 124.

In what has preceded (apart from the last paragraph) we have assumed simply that $$\phi(x)$$ has a derivative for all values of $$x$$ in the interval under consideration. If this condition is not fulfilled the theorems cease to be true. Thus Theorem B fails in the case of the function $y = 1 – \sqrt{x^{2}},$ where the square root is to be taken positive. The graph of this function is shown in Fig. 40. Here $$\phi(-1) = \phi(1) = 0$$: but $$\phi'(x)$$, as is evident from the figure, is equal to $$1$$ if $$x$$ is negative and to $$-1$$ if $$x$$ is positive, and never vanishes. There is no derivative for $$x = 0$$, and no tangent to the graph at $$P$$. And in this case $$x = 0$$ obviously gives a maximum of $$\phi(x)$$, but $$\phi'(0)$$, as it does not exist, cannot be equal to zero, so that the test for a maximum fails.

The bare existence of the derivative $$\phi'(x)$$, however, is all that we have assumed. And there is one assumption in particular that we have not made, and that is that $$\phi'(x)$$ itself is a continuous function. This raises a rather subtle but still a very interesting point. Can a function $$\phi(x)$$ have a derivative for all values of $$x$$ which is not itself continuous? In other words can a curve have a tangent at every point, and yet the direction of the tangent not vary continuously? The reader, if he considers what the question means and tries to answer it in the light of common sense, will probably incline to the answer No. It is, however, not difficult to see that this answer is wrong.

Consider the function $$\phi(x)$$ defined, when $$x \neq 0$$, by the equation $\phi(x) = x^{2}\sin(1/x);$ and suppose that $$\phi(0) = 0$$. Then $$\phi(x)$$ is continuous for all values of $$x$$. If $$x \neq 0$$ then $\phi'(x) = 2x \sin(1/x) – \cos(1/x);$ while $\phi'(0) = \lim_{h \to 0} \frac{h^{2}\sin(1/h)}{h} = 0.$ Thus $$\phi'(x)$$ exists for all values of $$x$$. But $$\phi'(x)$$ is discontinuous for $$x = 0$$; for $$2x\sin(1/x)$$ tends to $$0$$ as $$x \to 0$$, and $$\cos(1/x)$$ oscillates between the limits of indetermination $$-1$$ and $$1$$, so that $$\phi'(x)$$ oscillates between the same limits.

What is practically the same example enables us also to illustrate the point referred to at the end of § 121. Let $\phi(x) = x^{2}\sin(1/x) + ax,$ where $$0 < a < 1$$, when $$x \neq 0$$, and $$\phi(0) = 0$$. Then $$\phi'(0) = a > 0$$. Thus the conditions of Theorem A of § 121 are satisfied. But if $$x \neq 0$$ then $\phi'(x) = 2x\sin(1/x) – \cos(1/x) + a,$ which oscillates between the limits of indetermination $$a – 1$$ and $$a + 1$$ as $$x \to 0$$. As $$a – 1 < 0$$, we can find values of $$x$$, as near to $$0$$ as we like, for which $$\phi'(x) < 0$$; and it is therefore impossible to find any interval, including $$x = 0$$, throughout which $$\phi(x)$$ is a steadily increasing function of $$x$$.

It is, however, impossible that $$\phi'(x)$$ should have what was called in Ch. V (Ex. XXXVII. 18) a ‘simple’ discontinuity; that $$\phi'(x) \to a$$ when $$x \to +0$$, $$\phi'(x) \to b$$ when $$x \to -0$$, and $$\phi'(0) = c$$, unless $$a = b = c$$, in which case $$\phi'(x)$$ is continuous for $$x = 0$$. For a proof see § 125, Ex. XLVII. 3.

Example XLVI

1. Verify Theorem B when $$\phi(x) = (x – a)^{m} (x – b)^{n}$$ or $$\phi(x) = (x – a)^{m} (x – b)^{n} (x – c)^{p}$$, where $$m$$, $$n$$, $$p$$ are positive integers and $$a < b < c$$.

[The first function vanishes for $$x = a$$ and $$x = b$$. And $\phi'(x) = (x – a)^{m-1} (x – b)^{n-1} \{(m + n)x – mb – na\}$ vanishes for $$x = (mb + na)/(m + n)$$, which lies between $$a$$ and $$b$$. In the second case we have to verify that the quadratic equation $(m + n + p)x^{2} – \{m(b + c) + n(c + a) + p(a + b)\}x + mbc + nca + pab = 0$ has roots between $$a$$ and $$b$$ and between $$b$$ and $$c$$.]

2. Show that the polynomials $2x^{3} + 3x^{2} – 12x + 7,\quad 3x^{4} + 8x^{3} – 6x^{2} – 24x + 19$ are positive when $$x > 1$$.

3. Show that $$x – \sin x$$ is an increasing function throughout any interval of values of $$x$$, and that $$\tan x – x$$ increases as $$x$$ increases from $$-\frac{1}{2}\pi$$ to $$\frac{1}{2}\pi$$. For what values of $$a$$ is $$ax – \sin x$$ a steadily increasing or decreasing function of $$x$$?

4. Show that $$\tan x – x$$ also increases from $$x = \frac{1}{2}\pi$$ to $$x = \frac{3}{2}\pi$$, from $$x = \frac{3}{2}\pi$$ to $$x = \frac{5}{2}\pi$$, and so on, and deduce that there is one and only one root of the equation $$\tan x = x$$ in each of these intervals (cf. Ex. XVII. 4).

5. Deduce from Ex. 3 that $$\sin x – x < 0$$ if $$x > 0$$, from this that $$\cos x – 1 + \frac{1}{2}x^{2} > 0$$, and from this that $$\sin x – x + \frac{1}{6} x^{3} > 0$$. And, generally, prove that if \begin{aligned} C_{2m} & = \cos x – 1 + \frac{x^{2}}{2!} – \dots – (-1)^{m} \frac{x^{2m}}{{(2m)!}},\\ S_{2m+1}& = \sin x – x + \frac{x^{3}}{3!} – \dots – (-1)^{m} \frac{x^{2m+1}}{(2m+1)!},\end{aligned} and $$x> 0$$, then $$C_{2m}$$ and $$S_{2m+1}$$ are positive or negative according as $$m$$ is odd or even.

6. If $$f(x)$$ and $$f”(x)$$ are continuous and have the same sign at every point of an interval $${[a, b]}$$, then this interval can include at most one root of either of the equations $$f(x) = 0$$, $$f'(x) = 0$$.

7. The functions $$u$$, $$v$$ and their derivatives $$u’$$, $$v’$$ are continuous throughout a certain interval of values of $$x$$, and $$uv’ – u’v$$ never vanishes at any point of the interval. Show that between any two roots of $$u = 0$$ lies one of $$v = 0$$, and conversely. Verify the theorem when $$u = \cos x$$, $$v = \sin x$$.

[If $$v$$ does not vanish between two roots of $$u = 0$$, say $$\alpha$$ and $$\beta$$, then the function $$u/v$$ is continuous throughout the interval $${[\alpha, \beta]}$$ and vanishes at its extremities. Hence $$(u/v)’ = (u’v – uv’)/v^{2}$$ must vanish between $$\alpha$$ and $$\beta$$, which contradicts our hypothesis.]

8. Determine the maxima and minima (if any) of $$(x – 1)^{2} (x + 2)$$, $$x^{3} – 3x$$, $$2x^{3} – 3x^{2} – 36x + 10$$, $$4x^{3} – 18x^{2} + 27x – 7$$, $$3x^{4} – 4x^{3} + 1$$, $$x^{5} – 15x^{3} + 3$$. In each case sketch the form of the graph of the function.

[Consider the last function, for example. Here $$\phi'(x) = 5x^{2} (x^{2} – 9)$$, which vanishes for $$x = -3$$, $$x = 0$$, and $$x = 3$$. It is easy to see that $$x = -3$$ gives a maximum and $$x = 3$$ a minimum, while $$x = 0$$ gives neither, as $$\phi'(x)$$ is negative on both sides of $$x = 0$$.]

9. Discuss the maxima and minima of the function $$(x – a)^{m} (x – b)^{n}$$, where $$m$$ and $$n$$ are any positive integers, considering the different cases which occur according as $$m$$ and $$n$$ are odd or even. Sketch the graph of the function.

10. Discuss similarly the function $$(x – a) (x – b)^{2} (x – c)^{3}$$, distinguishing the different forms of the graph which correspond to different hypotheses as to the relative magnitudes of $$a$$, $$b$$, $$c$$.

11. Show that $$(ax + b)/(cx + d)$$ has no maxima or minima, whatever values $$a$$, $$b$$, $$c$$, $$d$$ may have. Draw a graph of the function.

12. Discuss the maxima and minima of the function $y = (ax^{2} + 2bx + c)/(Ax^{2} + 2Bx + {C}),$ when the denominator has complex roots.

[We may suppose $$a$$ and $$A$$ positive. The derivative vanishes if $\begin{equation*} (ax + b)(Bx + C) – (Ax + B)(bx + c) = 0. \tag{1} \end{equation*}$ This equation must have real roots. For if not the derivative would always have the same sign, and this is impossible, since $$y$$ is continuous for all values of $$x$$, and $$y \to a/A$$ as $$x \to +\infty$$ or $$x \to -\infty$$. It is easy to verify that the curve cuts the line $$y = a/A$$ in one and only one point, and that it lies above this line for large positive values of $$x$$, and below it for large negative values, or vice versa, according as $$b/a > B/A$$ or $$b/a < B/A$$. Thus the algebraically greater root of (1) gives a maximum if $$b/a > B/A$$, a minimum in the contrary case.]

13. The maximum and minimum values themselves are the values of $$\lambda$$ for which $$ax^{2} + 2bx + c – \lambda(Ax^{2} + 2Bx + C)$$ is a perfect square. [This is the condition that $$y = \lambda$$ should touch the curve.]

14. In general the maxima and maxima of $$R(x) = P(x)/Q(x)$$ are among the values of $$\lambda$$ obtained by expressing the condition that $$P(x) – \lambda Q(x) = 0$$ should have a pair of equal roots.

15. If $$Ax^{2} + 2Bx + C = 0$$ has real roots then it is convenient to proceed as follows. We have $y – (a/A) = (\lambda x + \mu)/\{A(Ax^{2} + 2Bx + C)\},$ where $$\lambda = bA – aB$$, $$\mu = cA – aC$$. Writing further $$\xi$$ for $$\lambda x + \mu$$ and $$\eta$$ for $$(A/\lambda^{2})(Ay – a)$$, we obtain an equation of the form $\eta = \xi/\{(\xi – p)(\xi – q)\}.$

This transformation from $$(x, y)$$ to $$(\xi, \eta)$$ amounts only to a shifting of the origin, keeping the axes parallel to themselves, a change of scale along each axis, and (if $$\lambda < 0$$) a reversal in direction of the axis of abscissae; and so a minimum of $$y$$, considered as a function of $$x$$, corresponds to a minimum of $$\eta$$ considered as a function of $$\xi$$, and vice versa, and similarly for a maximum.

The derivative of $$\eta$$ with respect to $$\xi$$ vanishes if $(\xi – p)(\xi – q) – \xi(\xi – p) – \xi(\xi – q) = 0,$ or if $$\xi^{2} = pq$$. Thus there are two roots of the derivative if $$p$$ and $$q$$ have the same sign, none if they have opposite signs. In the latter case the form of the graph of $$\eta$$ is as shown in Fig. 41a.

When $$p$$ and $$q$$ are positive the general form of the graph is as shown in Fig 41b, and it is easy to see that $$\xi = \sqrt{pq}$$ gives a maximum and $$\xi = -\sqrt{pq}$$ a minimum.2

In the particular case in which $$p = q$$ the function is $\eta = \xi/(\xi – p)^{2},$ and its graph is of the form shown in Fig. 41c.

The preceding discussion fails if $$\lambda = 0$$, if $$a/A = b/B$$. But in this case we have \begin{aligned} y – (a/A) &= \mu/\{A(Ax^{2} + 2Bx + C)\}\\ &= \mu/\{A^{2}(x – x_{1})(x – x_{2})\},\end{aligned} say, and $$dy/dx = 0$$ gives the single value $$x = \frac{1}{2}(x_{1} + x_{2})$$. On drawing a graph it becomes clear that this value gives a maximum or minimum according as $$\mu$$ is positive or negative. The graph shown in Fig. 42 corresponds to the former case.

[A full discussion of the general function $$y = (ax^{2} + 2bx + c)/(Ax^{2} + 2Bx + C)$$, by purely algebraical methods, will be found in Chrystal’s Algebra, vol i, pp. 464–7.]

16. Show that $$(x – \alpha)(x – \beta)/(x – \gamma)$$ assumes all real values as $$x$$ varies, if $$\gamma$$ lies between $$\alpha$$ and $$\beta$$, and otherwise assumes all values except those included in an interval of length $$4\sqrt{|\alpha – \gamma||\beta – \gamma|}$$.

17. Show that $y = \frac{x^{2} + 2x + c}{x^{2} + 4x + 3c}$ can assume any real value if $$0 < c < 1$$, and draw a graph of the function in this case.

18. Determine the function of the form $$(ax^{2} + 2bx + c)/(Ax^{2} + 2Bx + C)$$ which has turning values ( maxima or minima) $$2$$ and $$3$$ when $$x = 1$$ and $$x = -1$$ respectively, and has the value $$2.5$$ when $$x = 0$$.

19. The maximum and minimum of $$(x + a) (x + b)/(x – a) (x – b)$$, where $$a$$ and $$b$$ are positive, are $-\left(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} – \sqrt{b}}\right)^{2},\quad -\left(\frac{\sqrt{a} – \sqrt{b}}{\sqrt{a} + \sqrt{b}}\right)^{2}.$

20. The maximum value of $$(x – 1)^{2}/(x + 1)^{3}$$ is $$\frac{2}{27}$$.

21. Discuss the maxima and minima of $\begin{gathered} x(x – 1)/(x^{2} + 3x + 3),\quad x^{4}/(x – 1)(x – 3)^{3},\\ (x – 1)^{2}(3x^{2} – 2x – 37)/(x + 5)^{2}(3x^{2} – 14x – 1).\end{gathered}$

[If the last function be denoted by $$P(x)/Q(x)$$, it will be found that $P’Q – PQ’ = 72(x – 7)(x – 3)(x – 1)(x + 1)(x + 2)(x + 5).]$

22. Find the maxima and minima of $$a\cos x + b\sin x$$. Verify the result by expressing the function in the form $$A\cos(x – a)$$.

23. Find the maxima and minima of $a^{2}\cos^{2} x + b^{2}\sin^{2} x,\quad A\cos^{2}x + 2H\cos x\sin x + B\sin^{2} x.$

24. Show that $$\sin(x + a)/\sin(x + b)$$ has no maxima or minima. Draw a graph of the function.

25. Show that the function $\frac{\sin^{2}x}{\sin(x + a)\sin(x + b)}\quad (0 < a < b < \pi)$ has an infinity of minima equal to $$0$$ and of maxima equal to $-4\sin a\sin b/\sin^{2}(a – b).$

26. The least value of $$a^{2}\sec^{2}x + b^{2}\csc^{2}x$$ is $$(a + b)^{2}$$.

27. Show that $$\tan 3x \cot 2x$$ cannot lie between $$\frac{1}{9}$$ and $$\frac{3}{2}$$.

28. Show that, if the sum of the lengths of the hypothenuse and another side of a right-angled triangle is given, then the area of the triangle is a maximum when the angle between those sides is $$60^\circ$$.

29. A line is drawn through a fixed point $$(a, b)$$ to meet the axes $$OX$$, $$OY$$ in $$P$$ and $$Q$$. Show that the minimum values of $$PQ$$, $$OP + OQ$$, and $$OP\cdot OQ$$ are respectively $$(a^{2/3} + b^{2/3})^{3/2}$$, $$(\sqrt{a} + \sqrt{b})^{2}$$, and $$4ab$$.

30. A tangent to an ellipse meets the axes in $$P$$ and $$Q$$. Show that the least value of $$PQ$$ is equal to the sum of the semiaxes of the ellipse.

31. Find the lengths and directions of the axes of the conic $ax^{2} + 2hxy + by^{2} = 1.$

[The length $$r$$ of the semi-diameter which makes an angle $$\theta$$ with the axis of $$x$$ is given by $1/r^{2} = a\cos^{2} \theta + 2h\cos\theta \sin\theta + b\sin^{2} \theta.$ The condition for a maximum or minimum value of $$r$$ is $$\tan 2\theta = 2h/(a – b)$$. Eliminating $$\theta$$ between these two equations we find $\{a – (1/r^{2})\} \{b – (1/r^{2})\} = h^{2}.]$

32. The greatest value of $$x^{m}y^{n}$$, where $$x$$ and $$y$$ are positive and $$x + y = k$$, is $m^{m} n^{n} k^{m+n}/(m + n)^{m+n}.$

33. The greatest value of $$ax + by$$, where $$x$$ and $$y$$ are positive and $$x^{2} + xy + y^{2} = 3\kappa^{2}$$, is $2\kappa \sqrt{a^{2} – ab + b^{2}}.$

[If $$ax + by$$ is a maximum then $$a + b(dy/dx) = 0$$. The relation between $$x$$ and $$y$$ gives $$(2x + y) + (x + 2y)(dy/dx) = 0$$. Equate the two values of $$dy/dx$$.]

34. If $$\theta$$ and $$\phi$$ are acute angles connected by the relation $$a \sec\theta + b \sec\phi = c$$, where $$a$$, $$b$$, $$c$$ are positive, then $$a\cos\theta + b\cos\phi$$ is a minimum when $$\theta = \phi$$.

1. A function which is continuous but has no derivative may have maxima and minima. We are of course assuming the existence of the derivative.↩︎
2. The maximum is $$-1/(\sqrt{p} – \sqrt{q})^{2}$$, the minimum $$-1/(\sqrt{p} + \sqrt{q})^{2}$$, of which the latter is the greater.↩︎