Julian's musings

Here’s the graph of $y=-\dfrac{1}{x}$ for $x\ne0$.

Where is this function increasing? Is it an increasing function?

Looking at various recent examination papers, it has become clear to me that there is significant confusion between these two questions. This post is intended to bring some clarity to the situation.

At the start of this post, I will give an example of the confusion as it appears in exam questions (and probably elsewhere), and clarify what the two different phrases mean using the above example. I will then delve more deeply into the mathematics of these two things, going beyond A-level content, and use some undergraduate analysis to find equivalent conditions for them in terms of the derivatives of the functions. It is fine to skip over the technical stuff and just look at the results (theorems)!

(Exactly the same applies to the use of the term “decreasing”, but for simplicity we will focus on increasing functions in this post.)

Here is an example of a question (based on a real exam question) which typifies the confusion.

The equation of a curve is $y=x^3+4x^2-5x$.

Find the set of values of $x$ for which $y$ is an increasing function of $x$.

If we replace “increasing function” by another familiar A-level term describing functions, “one-to-one function”, the question becomes:

A function is given by $f(x)=x^3+4x^2-5x$.

Find the set of values of $x$ for which $f(x)$ is a one-to-one function of $x$.

This is clearly nonsensical, because whether a function is one-to-one or not is a property of the function as a whole, not a property of the function values at any particular input value.

Likewise, a function either is or is not an increasing function; it is a property of the function as a whole.

Informally (and not quite correctly), we can describe the difference as follows:

• A function is an increasing function if larger input values give larger output values.
• A function is increasing at a point if at that point, the function has a positive gradient.

An example which shows that these are not the same is the function $f(x)=-\dfrac{1}{x}$ for $x\ne0$ shown above. This function is increasing at every value of $x\ne0$, as the gradient is always positive. However, it is not an increasing function, because $f(1)<f(-1)$. If, though, we restricted the domain of the function to $x>0$, then it would be an increasing function.

So the above-quoted exam question does not make any sense, just as the modified version did not: either $y$ is an increasing function of $x$ or it is not. If the question had instead asked “Find the set of values of $x$ at which $y$ is increasing,” it would have been fine.

Incidentally, the idea of increasing and decreasing functions connects very well with the issue of rearranging inequalities (increasing the depth of connections within the subject): a function can be applied to both sides of an inequality without changing the direction of the inequality if the function is (strictly) increasing; it can be applied but with a change in the direction of the inequality if the function is (strictly) decreasing, and if the function is neither, then the function cannot be applied to the inequality. So we cannot square both sides of an inequality unless we are restricted to non-negative values, and we cannot take the reciprocal of an inequality unless we have the same restriction (and in that case, we must also reverse the direction of the inequality).

It seems reasonable to assert that if a function is an increasing function, then it will be increasing at every point. There turns out to be some subtlety to this, as we now delve into a little more deeply.

A formal definition of increasing

We can give a formal definition of an increasing function. For example, this definition is from Apostol, Mathematical Analysis, 2nd ed, p94, and identical definitions appear on the internet:

Definition 1: Let $f$ be a real-valued function whose domain is a subset $S$ of $\mathbb{R}$. Then $f$ is said to be an increasing (or nondecreasing) function if for every pair of points $x$ and $y$ in $S$, $x<y$ implies $f(x)\le f(y)$. If $x<y$ implies $f(x)<f(y)$, then $f$ is said to be a strictly increasing function. (Decreasing functions are similarly defined.)

Note the distinction between increasing and strictly increasing here: a constant function such as $f(x)=0$ for $x\in\mathbb{R}$ is both an increasing and decreasing function, though it is not a strictly increasing function.

We could also try to come up with a definition of increasing at a point. There are no standard definitions of this idea, and the following proposed definition is certainly beyond A-level in its formality. It is based on the definition of continuity, which is about the behaviour of a function “near” to a point.

Definition 2: Let $f$ be a real-valued function whose domain is a subset $S$ of $\mathbb{R}$. Then $f$ is said to be increasing at the point $x$ in $S$ if there is some $\delta>0$ such that:

for every $y$ in $S$ with $x<y<x+\delta$, $f(x)\le f(y)$, and for every $y$ in $S$ with $x-\delta<y<x$, $f(y)\le f(x)$.

If the $\le$ signs are replaced by $<$ signs in these two inequalities, then $f$ is said to be strictly increasing at $x$.

With this definition, the above exam question (reworded) makes sense, and the correct final answer is what the examiner would expect. (One might wonder whether one could make such a local definition of one-to-one, and indeed, this is done when considering the Inverse Function and Implicit Function theorems. But that is a story for another day.)

Using calculus

So far, no calculus has appeared, yet we typically teach our students to determine whether a function is an increasing function or to find where it is increasing by differentiating the function. So let us now consider how we could use calculus to help us.

For us to be able to use calculus, we need to assume that our function is differentiable throughout $S$. We could then propose the following theorem:

Theorem 1 (incorrect attempt): Let $f$ be a real-valued continuous function whose domain is a subset $S$ of $\mathbb{R}$ and is differentiable at every (interior) point of $S$. Then $f$ is an increasing function if and only if $f’(x)>0$ for all $x$ in (the interior of) $S$.

(The use of “interior” is to avoid certain technical complications.)

Unfortunately this fails immediately: the constant function $f(x)=0$ for $x\in\mathbb{R}$ is increasing, yet $f’(x)=0$.

We could try changing this to say that $f$ is a strictly increasing function, but that fails if the function has a point of inflection. For example, $f(x)=x^3$ is a strictly increasing function, even though its derivative is zero at $x=0$.

We could also try changing the condition to say that $f’(x)\ge0$ for all $x$ in $S$. However, this also fails: if the graph has a discontinuity, such as the function $f(x)=-\dfrac{1}{x}$ for $x\ne0$ that we looked at before, then it might have $f’(x)>0$ for all $x$ in $S$, yet not be an increasing function.

This feels more hopeful, though: after all, the only problem now is the “hole” in the domain $S$. And it turns out that if we restrict the domain to be an interval (that is, a subset of the reals with no “holes”), then it will work:

Theorem 1 (correct version): Let $f$ be a real-valued continuous function whose domain is an interval $I$ of $\mathbb{R}$ and is differentiable at every point in (the interior of) $I$. Then $f$ is an increasing function if and only if $f’(x)\ge 0$ for all $x$ in (the interior of) $I$.

The formal proof of this is found below, and though it is quite technical, the theorem itself seems clearly true, and school students could probably be convinced to believe it (at least once it is written in more student-friendly language).

What can we say, though, about whether a (differentiable) function is increasing at a point? Using Definition 2 above, we get the corresponding theorem:

Theorem 2: Let $f$ be a real-valued continuous function whose domain is an interval $I$ of $\mathbb{R}$ and which is differentiable at every (interior) point of $I$. Then is $f$ is increasing at the point $x$ in $I$ if and only if there is some $\delta>0$ for which $f’(y)\ge0$ for all $y$ in (the interior of) $I$ with $x-\delta<y<x+\delta$.

Why is it not sufficient to just require $f’(x)\ge0$? Well, consider the functions $f(x)=x^3$ and $f(x)=-x^3$. They both have $f’(x)=0$, yet the first is increasing (indeed, even strictly increasing) at $x=0$, while the second is decreasing at $x=0$. And a function such as $f(x)=x^2$ is neither increasing nor decreasing at $x=0$. So we really do need to consider a small interval around the point of interest.

(Theorem 2 could be extended, with care, to more general subsets of $\mathbb{R}$, as we are only discussing a local property of the function. But it is not particularly interesting to do so.)

So the question of determining at which points a function is increasing (or decreasing) is more subtle than it appears: not only does one have to find where the function has derivative $\le0$ (and not just $<0$), but one also has to determine what is happening at those points where the derivative is zero, as there are different types of stationary points. (At those points where the derivative is strictly positive, the function is certainly strictly increasing, which follows from Theorem 4 below.)

Things get more complicated if we now wish to consider strictly increasing (or decreasing) functions. There is a relatively weak theorem which will suffice much of the time:

Theorem 3: Let $f$ be a continuous real-valued function whose domain is an interval $I$ of $\mathbb{R}$ and which is differentiable at every (interior) point of $I$. Then if $f’(x)>0$ throughout $I$, $f$ is a strictly increasing function.

Note that this is a one-directional theorem; $f(x)=x^3$ for $x\in\mathbb{R}$ is our standard example of a strictly increasing function which does not have $f’(x)>0$ throughout the domain because of the point of inflection at the origin. The proof of Theorem 3 follows exactly as that of Theorem 1.

An easy corollary of this is the following (local) theorem:

Theorem 4: Let $f$ be a continuous real-valued function whose domain is a subset $S$ of $\mathbb{R}$. If $f$ is differentiable at the point $x$ in the interior of $S$ and $f’(x)>0$, then $f$ is strictly increasing at $x$.

This is the theorem which is typically used when answering A-level exam questions such as the one above. Unfortunately, as we see from our example of $f(x)=x^3$, this too is a one-directional theorem: every point at which $f’(x)>0$ is a point at which the function is strictly increasing, but there may be other points where this is the case but where $f’(x)=0$. (If $f’(x)<0$, then the function is strictly decreasing at this point, so it cannot be increasing.) The question of using calculus to determine where a function is increasing, rather than strictly increasing, is somewhat more complicated, as we see from Theorem 2 above. But at A-level, the functions are always nice enough that the only difficulties will be at the stationary points.

There is actually a necessary and sufficient condition for a function to be strictly increasing, but this is more subtle:

Theorem 5: Let $f$ be a continuous real-valued function whose domain is an interval $I$ of $\mathbb{R}$ and which is differentiable at every interior point of $I$. Then $f$ is strictly increasing on $I$ if and only if $f’(x)\ge0$ throughout $I$ and there is no non-trivial subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the interior of $J$.

The proof can be found below.

Teaching this topic at A-level

Putting this all together, we see that Theorem 4 is the crucial theorem for school use. Teaching the meaning of the term “increasing function” (Definition 1) and a simplified explanation of “increasing at a point” (Definition 2), along with Theorem 4 should give a good grounding. It would also be wise to caution that it is a one-way theorem by comparing and contrasting examples such as $f(x)=x^2$ and $f(x)=x^3$.

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Proofs of Theorems 1 and 5

This technical appendix uses tools from undergraduate analysis. The proofs of the other three theorems are very similar to these or they follow immediately from these.

Theorem 1

Let $f$ be a real-valued continuous function whose domain is an interval $I$ of $\mathbb{R}$ and is differentiable at every point in the interior of $I$. Then $f$ is an increasing function if and only if $f’(x)\ge 0$ for all $x$ in the interior of $I$.

Proof

We show first that if $f$ is an increasing function, then $f’(x)\ge0$ for all $x$ in the interior of $I$, and we argue by contradiction. Assume that $f’(x_0)<0$ for some $x_0$ in the interior of $I$. Using the definition of derivative, this means that $\lim\limits_{\substack{x\to x_0\\ x\in I}}\dfrac{f(x)-f(x_0)}{x-x_0}<0$. So there is some $x_1\in I$ (where $x_1\ne x_0$) with $\dfrac{f(x)-f(x_0)}{x-x_0}<0$ (otherwise the limit would be $\ge0$). If $x_1>x_0$, then multiplying by $x_1-x_0$ gives $f(x_1)-f(x_0)<0$, so $f(x_1)<f(x_0)$, If $x_1<x_0$, then multiplying by $x_1-x_0$ gives $f(x_1)-f(x_0)>0$, so $f(x_1)>f(x_0)$. Either way, this shows that the function is not increasing on $I$, and we have our desired contradition. Thus if $f$ is an increasing function, we must have $f’(x)\ge0$ for all $x$ in the interior of $I$.

Conversely, if $f’(x)\ge0$ for all $x$ in the interior of $I$, then let $x<y$ be any two points in $I$. Then by the mean-value theorem, there is some $z$ with $x<z<y$ for which $f(y)-f(x)=f’(z)(y-x)$ (and note that $z$ lies in the interior of $I$ as $I$ is an interval). Since $f’(z)\ge0$ by assumption, and $y-x>0$, it follows that $f(y)-f(x)\ge0$, so $f(x)\le f(y)$. Therefore $f$ is an increasing function.

Theorem 5

Let $f$ be a continuous real-valued function whose domain is an interval $I$ of $\mathbb{R}$ and which is differentiable at every interior point of $I$. Then $f$ is strictly increasing on $I$ if and only if $f’(x)\ge0$ throughout $I$ and there is no non-trivial subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the interior of $J$.

Proof

We first prove that if the derivative condition is not met, then $f$ is not strictly increasing on $I$. If $f’(x)<0$ at any point in $I$, then $f$ is not increasing (by Theorem 1), so it is certainly not strictly increasing. If $f’(x)\ge0$ throughout $I$ but there is a non-trivial subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the interior of $J$, then $f$ is constant throughout $J$ (by the mean-value theorem). In particular, there are $y<z$ in $J$ with $f(y)=f(z)$, showing that $f$ is not strictly increasing.

Conversely, if $f’(x)\ge0$ throughout $I$, then $f$ is increasing by Theorem 1. Assume now that there is no non-trivial subinterval $J$ of $I$ with $f’(x)=0$ for all $x$ in the interior of $J$. But if $f$ were not strictly increasing, then there would be $y<z$ in $I$ with $f(y)=f(z)$, so $f(x)$ is constant on the interval $y<x<z$. (For if $f(y)<f(x)$ for some $x$ in this interval, we would have $f(x)>f(z)$, contradicting $f$ increasing.) Therefore $f’(x)=0$ throughout this interval, contradicting our assumption. So $f$ must be strictly increasing.