The Increasing Function Theorem (IFT), which connects positive derivatives and increasing functions, is absolutely foundational to a first course in calculus: for example, it justifies the use of the first derivative test for extreme values. The theorem feels intuitively obvious (positive slope means increasing, right?) but proving it is harder than one might expect. For this reason, many books “prove” the IFT using the Mean Value Theorem (MVT). But this can hardly be called rigorous if the MVT is left unproved? Even worse is to prove the MVT using the Extreme Value Theorem, but then leave that unproved. If we’re going to tell students to be suspicious of their intuition, we need to give them something better in its place.
One can simply say to students “the IFT is true, but its proof is harder than you might expect. It is not easy to relate the properties of a function across a whole interval to its properties at individual points in that interval.” Then there is no need to muddy the waters using unproved theorems to justify a claim that students probably found intuitive in the first place.
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Suppose you want to prove the IFT. Maybe just for one interested student, or for a class of students getting ready to be math majors. You don’t want to give a full course on analysis, but you still want to give a good feel for why the theorem is true. If so, the proof below is a good option. Without invoking any existence theorems, this proof shows how to construct a point at which the derivative is positive. Also, the proof has a relatively clear idea behind it: it gives students a clear sense for how “increasing on an interval” relates to “increasing at a point,” and provides a model for thinking about the relationships between the local and global properties of a function.
The idea of this proof is illustrated in the figure below. By successively halving the interval and always choosing the half with the greater (or equal) slope, we gradually hone in on a specific point at which the slope must be greater than or equal to the average slope on the interval (thus, if the average slope is ever positive, there must be a point at which the derivative is positive). Below, I fill in the details needed to turn this idea into a proof. But even without the details, sketching graphs like the ones below can give students an extremely good intuitive sense of why the theorem is true and what type of reasoning is needed for this sort of proof.
The proof below uses the nested interval property (which is an axiom of the real numbers) and the fact that, if f is differentiable, then any 2-sided derivative at a given point will equal the true derivative at the given point1. This fact is relatively easy to prove, and an excellent exercise for anyone who is advanced enough to be thinking about a proof of the IFT.
We are given an interval [a,b] and a function that is differentiable on this interval. We will construct a point at which \(\scriptsize f^\prime(x)\geq \frac{f(b)-f(a)}{b-a}\) . The existence of such a point implies that, if \(\scriptsize f(b)\geq f(a)\) then there must be a point where the derivative is zero or positive. Similarly, if \(\scriptsize f(b)\leq f(a)\), then there must be a point where the derivative is zero or negative. The absence of any nonpositive values of \(\scriptsize f^\prime\) thus implies that the function is strictly increasing.
Let \(\scriptsize x_1\) be the midpoint of \(\scriptsize [a,b]\), i.e. \(\scriptsize x_1=\frac{a+b}{2}\). We first note that the average slope of \(\scriptsize f\) on \(\scriptsize [a,b]\) equals the average of its two average slopes on \(\scriptsize [a,x_1]\) and \(\scriptsize [x_1,b]\), i.e. that
$$\scriptsize \frac{f(b)-f(a)}{b-a}=\frac{\frac{f(b)-f(x_1)}{b-x_1}+\frac{f(x_1)-f(a)}{x_1-a}}{2} .$$
This equality comes from substituting \(\scriptsize \frac{b-a}{2}\) for both \(\scriptsize (b-x_1)\) and \(\scriptsize (x_1-a)\) in the right hand side of the above equation, which we can do since \(\scriptsize x_1\) is the mindpoint of \(\scriptsize [a,b]\).
Since \(\scriptsize \frac{f(b)-f(a)}{b-a}\) is the average of \(\scriptsize \frac{f(b)-f(x_1)}{b-x_1}\) and \(\scriptsize \frac{f(x_1)-f(a)}{x_1-a}\), the average slope on at least one of the intervals \(\scriptsize [a,x_1]\) or \(\scriptsize [x_1,b]\) must be greater than or equal to the average slope on the full interval \(\scriptsize [a,b]\). Pick a sub-interval (either \(\scriptsize [a,x_1]\) or \(\scriptsize [x_1,b]\)) on which the average slope is greater than or equal to the average slope on \(\scriptsize [a,b]\), and call this interval \(\scriptsize I_1\).
In a similar fashion, we may now divide \(\scriptsize I_1\) in half, and in one of the two halves (call it \(\scriptsize I_2\)) the average slope must be greater than or equal to the average slope in all of \(\scriptsize I_1\), which in turn is greater than or equal to the average slope on \(\scriptsize [a,b]\).
We now have an interval \(\scriptsize I_2\), which is a quarter the size of the original interval \(\scriptsize [a,b]\) and for which the average slope is greater than or equal to the average slope on \(\scriptsize [a,b]\). Continuing in this manner, we split \(\scriptsize I_2\) in half, and thereby produce an interval \(\scriptsize I_3\) whose length is an eighth that of \(\scriptsize [a,b]\) and on which the average slope is greater than or equal to the average slope on \(\scriptsize [a,b]\).
This process results in nested intervals \(\scriptsize I_n\), in each of which the average slope of \(\scriptsize f\) is greater than or equal to its average slope on \(\scriptsize [a,b]\). Since these nested intervals have lengths approaching zero, the nested interval theorem guarantees the existence of a single point contained in all \(\scriptsize I_n\); we call this point \(\scriptsize c_1\). The two-sided derivative theorem then guarantees that, as \(\scriptsize n\) gets large, the average slope of \(\scriptsize f\) on the interval \(\scriptsize I_n\) gets arbitrarily close to \(\scriptsize f^\prime (c_1)\).
We can therefore conclude that \(\scriptsize f^\prime (c_1)\geq\frac{f(b)-f(a)}{b-a}\), since there are numbers that are greater than or equal to \(\scriptsize \frac{f(b)-f(a)}{b-a}\) and are arbitrarily close to \(\scriptsize f^\prime (c_1)\)
Given two points on f with a positive slope between them, we can construct a point with a positive derivative; and given two points on f with a negative slope between them, we can construct a point with a negative derivative. The Increasing Function Theorem follows immediately from these constructions.
- If f is not differentiable, some two sided derivatives may converge anyway, and they may have different values. But when f is differentiable, all 2-sided derivatives will match the true derivative. ↩︎