In all that follows we suppose that $$\phi(x)$$ is a function of $$x$$ which has a derivative $$\phi'(x)$$ for all values of $$x$$ in question. This assumption of course involves the continuity of $$\phi(x)$$.

The meaning of the sign of $$\phi'(x)$$.

Theorem A. If $$\phi'(x_{0}) > 0$$ then $$\phi(x) < \phi(x_{0})$$ for all values of $$x$$ less than $$x_{0}$$ but sufficiently near to $$x_{0}$$, and $$\phi(x) > \phi(x_{0})$$ for all values of $$x$$ greater than $$x_{0}$$ but sufficiently near to $$x_{0}$$.

For $$\{\phi(x_{0} + h) – \phi(x_{0})\}/h$$ converges to a positive limit $$\phi'(x_{0})$$ as $$h \to 0$$. This can only be the case if $$\phi(x_{0} + h) – \phi(x_{0})$$ and $$h$$ have the same sign for sufficiently small values of $$h$$, and this is precisely what the theorem states. Of course from a geometrical point of view the result is intuitive, the inequality $$\phi'(x) > 0$$ expressing the fact that the tangent to the curve $$y = \phi(x)$$ makes a positive acute angle with the axis of $$x$$. The reader should formulate for himself the corresponding theorem for the case in which $$\phi'(x) < 0$$.

An immediate deduction from Theorem A is the following important theorem, generally known as Rolle’s Theorem. In view of the great importance of this theorem it may be well to repeat that its truth depends on the assumption of the existence of the derivative $$\phi'(x)$$ for all values of $$x$$ in question.

Theorem B. If $$\phi(a) = 0$$ and $$\phi(b) = 0$$, then there must be at least one value of $$x$$ which lies between $$a$$ and $$b$$ and for which $$\phi'(x) = 0$$.

There are two possibilities: the first is that $$\phi(x)$$ is equal to zero throughout the whole interval $${[a, b]}$$. In this case $$\phi'(x)$$ is also equal to zero throughout the interval. If on the other hand $$\phi(x)$$ is not always equal to zero, then there must be values of $$x$$ for which $$\phi(x)$$ is positive or negative. Let us suppose, for example, that $$\phi(x)$$ is sometimes positive. Then, by Theorem 2 of § 102, there is a value $$\xi$$ of $$x$$, not equal to $$a$$ or $$b$$, and such that $$\phi(\xi)$$ is at least as great as the value of $$\phi(x)$$ at any other point in the interval. And $$\phi'(\xi)$$ must be equal to zero. For if it were positive then $$\phi(x)$$ would, by Theorem A, be greater than $$\phi(\xi)$$ for values of $$x$$ greater than $$\xi$$ but sufficiently near to $$\xi$$, so that there would certainly be values of $$\phi(x)$$ greater than $$\phi(\xi)$$. Similarly we can show that $$\phi'(\xi)$$ cannot be negative.

Cor 1. If $$\phi(a) = \phi(b) = k$$, then there must be a value of $$x$$ between $$a$$ and $$b$$ such that $$\phi'(x) = 0$$.

We have only to put $$\phi(x) – k = \psi(x)$$ and apply Theorem B to $$\psi(x)$$.

Cor 2. If $$\phi'(x) > 0$$ for all values of $$x$$ in a certain interval, then $$\phi(x)$$ is an increasing function of $$x$$, in the stricter sense of § 95, throughout that interval.

Let $$x_{1}$$ and $$x_{2}$$ be two values of $$x$$ in the interval in question, and $$x_{1} < x_{2}$$. We have to show that $$\phi(x_{1}) < \phi(x_{2})$$. In the first place $$\phi(x_{1})$$ cannot be equal to $$\phi(x_{2})$$; for, if this were so, there would, by Theorem B, be a value of $$x$$ between $$x_{1}$$ and $$x_{2}$$ for which $$\phi'(x) = 0$$. Nor can $$\phi(x_{1})$$ be greater than $$\phi(x_{2})$$. For, since $$\phi'(x_{1})$$ is positive, $$\phi(x)$$ is, by Theorem A, greater than $$\phi(x_{1})$$ when $$x$$ is greater than $$x_{1}$$ and sufficiently near to $$x_{1}$$. It follows that there is a value $$x_{3}$$ of $$x$$ between $$x_{1}$$ and$$~x_{2}$$ such that $$\phi(x_{3}) = \phi(x_{1})$$; and so, by Theorem B, that there is a value of $$x$$ between $$x_{1}$$ and $$x_{3}$$ for which $$\phi'(x) = 0$$.

Cor 3. The conclusion of Cor. still holds if the interval $${[a, b]}$$ considered includes a finite number of exceptional values of $$x$$ for which $$\phi'(x)$$ does not exist, or is not positive, provided $$\phi(x)$$ is continuous even for these exceptional values of $$x$$.

It is plainly sufficient to consider the case in which there is one exceptional value of $$x$$ only, and that corresponding to an end of the interval, say to $$a$$. If $$a < x_{1} < x_{2} < b$$, we can choose $$a + \delta$$ so that $$a + \delta < x_{1}$$, and $$\phi'(x) > 0$$ throughout $${[a + \delta, b]}$$, so that $$\phi(x_{1}) < \phi(x_{2})$$, by Cor. 2. All that remains is to prove that $$\phi(a) < \phi(x_{1})$$. Now $$\phi(x_{1})$$ decreases steadily, and in the stricter sense, as $$x_{1}$$ decreases towards $$a$$, and so $\phi(a) = \phi(a + 0) = \lim_{x_{1}\to a+0} \phi(x_{1}) < \phi(x_{1}).$

Cor 4. If $$\phi'(x) > 0$$ throughout the interval $${[a, b]}$$, and $$\phi(a) \geq 0$$, then $$\phi(x)$$ is positive throughout the interval $${[a, b]}$$.

The reader should compare the second of these corollaries very carefully with Theorem A. If, as in Theorem A, we assume only that $$\phi'(x)$$ is positive at a single point $$x = x_{0}$$, then we can prove that $$\phi(x_{1}) < \phi(x_{2})$$ when $$x_{1}$$ and $$x_{2}$$ are sufficiently near to $$x_{0}$$ and $$x_{1} < x_{0} < x_{2}$$. For $$\phi(x_{1}) < \phi(x_{0})$$ and $$\phi(x_{2}) > \phi(x_{0})$$, by Theorem A. But this does not prove that there is any interval including $$x_{0}$$ throughout which $$\phi(x)$$ is a steadily increasing function, for the assumption that $$x_{1}$$ and $$x_{2}$$ lie on opposite sides of $$x_{0}$$ is essential to our conclusion. We shall return to this point, and illustrate it by an actual example, in a moment (§ 124).