Is $x^3$ really an increasing function for all intervals?

Question

I had an argument with my maths teacher today... He says, along with another classmate of mine that $x^3$ is increasing for all intervals. I argue that it isn't. If we look at conditions for increasing, decreasing and stable states of a function, for a function $f(x)$:

function is increasing for $f'(x) > 0$

function is decreasing for $f'(x) < 0$

(as per my knowledge) function is neither increasing nor decreasing (i.e. it's stable) for $f'(x) = 0$

so for $x = 0$, function $f(x) = x^3$ should be stable, NOT INCREASING. Hence the interval for which the function must be increasing, according to me, would be $\mathbb{R}$ - {$0$}.

This means, the function $f(x)=x^3$ should be increasing for all real values EXCEPT $x=0$.

My teacher denied, saying that we consider values of $f'(x) >= 0$. My response was to look at the function $f(x) = x^2$ which is increasing for interval $(0, \infty]$ and decreasing for interval $(\infty, 0]$, without a doubt.

again, I reason that this time too, $x=0$ is neither decreasing nor increasing hence it is not included in either of the intervals. My teacher, had nothing to say on this, simply telling me that these two examples are completely incomparable.

[another reasoning that my classmate and my teacher gave was that $x=0$ is a single point. Thus the function has to be constant there. Since the function is increasing before and after the point $x=0$, the function must be increasing for value $x=0$ as well.

In my opinion, that is completely against the very idea of calculus. When we define the rate of change at a point, it is solely attributed to that point alone. The limit of the rate of change, as points around $x=0$ close in on 0 is 0, hence in my opinion, the function should be constant at $x=0$, while increasing elsewhere in the set of real numbers]

Am I correct?

Answer 1

When we say $f$ is increasing over an interval $I$, we are not making a statement about $f’$ at all, we are making a statement about $f$, namely that for $a,b\in I$ with $a<b$ we have $f(a)<f(b)$.

Since this is true for all intervals for the function $f(x)=x^3$, we say the function is increasing on all intervals.

Note this definition applies for functions that are not even differentiable everywhere, so we certainly don’t need or want to use the derivative in our definition of when a function is increasing.

Remark

A $C^1$ function (i.e., one with a continuous derivative) satisfying $f’(a)>0$ will be guaranteed to be increasing on some interval containing $a$, but as your example of $x^3$ shows, the converse is not necessarily true.

Update

Having seen the discussion in the comments, I'm gathering that you have something of a philosophical objection to the above characterization of "increasing", and would prefer an infinitesimal definition based on the derivative, as more in keeping with "the very idea of calculus", as you put it in the main post.

The definition of "increasing" I've given is essentially standard (some would add the modifier "strictly" for clarity), but based on your concerns, let me also offer a sort of philosophical justification for why we define things the way we do.

While infinitesimal concepts (like the derivative) are certainly an important part of calculus, a perhaps even more important part of the subject concerns the relationship between infinitesimal information, and local and global behavior of a function.

This is famously seen in the fundamental theorem of calculus, which relates the infinitesimal rate of change with the global change over an interval, but is also seen in statements like "If $f$ has a local extremum at $x$, and $f$ is differentiable at $x$, then $f'(x)=0$" (note that the reverse is not true, as the example of $x^3$ shows).

In other words, while we obtain infinitesimal information from the derivative, we are often doing so not for its own sake, but rather in order to answer questions that are not about infinitesimal behavior, but instead about local and global behavior of a function. And whether $f$ is increasing or not is one such question.

Final clarification

[upgraded from comments since it relates to the original question]

I see near the end of your post you mentioned the example of $f(x)=x^2$. Here there is a subtle distinction at play. It is true that $f$ is increasing over $[0,\infty)$. However, when we say $f$ is increasing at a point, that's usually shorthand for "increasing over some open interval containing that point."

This creates two slightly different questions:

1. What's the largest interval over which $f$ is increasing? (answer: $[0,\infty)$).
2. What's the set of points at which $f$ is increasing? (answer: $(0,\infty)$).

However, the first question doesn't always have a well defined answer - consider something like $f(x)=2x^3-3x^2$, which is increasing on $(-\infty,0]$ and on $[1,\infty)$, but not over the union of those intervals, (since $f(1)<f(0)$), so which interval would you pick to be the answer? Therefore, usually we ask the second question instead (in this final example the answer would be $(-\infty,0)\cup(1,\infty)$).

Answer 2

Your teacher is correct. For a function $f:\mathbb R\to\mathbb R$ to be strictly increasing, we need $x<y$ to imply $f(x)<f(y)$. So let's see if this is the case for $f:\mathbb R\to\mathbb R$ given by $f(x)=x^3$.

If I understand you correctly, you are convinced that we have $f(x)<f(y)$ whenever $x<y<0$ and whenever $0<x<y$. So let us look at the other cases.

Firstly, if $x<0\leq y$, then $f(x)<0\leq f(y)$, and so in particular $f(x)<f(y)$. If $x\leq 0< y$, then $f(x)\leq 0<f(y)$, and so again, $f(x)<f(y)$.

In total, we end up with what we want: $f$ is indeed strictly increasing on $\mathbb R$.

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What you might want to keep in mind is the following: if $f:\mathbb R\to\mathbb R$ satisfies $f'(x)>0$ for all $x\in\mathbb R$, except for finitely many $x$ for which we have $f'(x)=0$, then $f$ is strictly increasing on $\mathbb R$. The proof of this statement is similar to our above discussion.

Answer 3

To some extent, this comes down to how you're going to define your terms. I would argue that the meaning of ‘increasing’ and ‘decreasing’ doesn't depend on Caclulus; we can explain it with a few inequalities, and even use it in contexts where we're not talking about anything like real numbers (so that Calculus doesn't even make sense there). I use the following four definitions:

• A function $ f $ is strictly increasing on a set $ S $ if, whenever $ x $ and $ y $ are in $ S $ and $ x < y $, then $ f ( x ) $ and $ f ( y ) $ exist and $ f ( x ) < f ( y ) $;
• A function $ f $ is weakly increasing on a set $ S $ if, whenever $ x $ and $ y $ are in $ S $ and $ x \leq y $, then $ f ( x ) $ and $ f ( y ) $ exist and $ f ( x ) \leq f ( y ) $;
• A function $ f $ is strictly decreasing on a set $ S $ if, whenever $ x $ and $ y $ are in $ S $ and $ x < y $, then $ f ( x ) $ and $ f ( y ) $ exist and $ f ( x ) > f ( y ) $;
• A function $ f $ is weakly decreasing on a set $ S $ if, whenever $ x $ and $ y $ are in $ S $ and $ x \leq y $, then $ f ( x ) $ and $ f ( y ) $ exist and $ f ( x ) \geq f ( y ) $.

Depending on context, one usually takes either ‘strictly’ or ‘weakly’ as the default, so that one of these adverbs doesn't need to be said; in your class, I think that you'd take ‘strictly’ as the default.

Notice that by these definitions, it makes no sense to say that a function is increasing at a point, only on a set (which is typically an open interval in Calculus).

Now when $ f $ is real-valued and $ S $ is an interval in the real line, we have these theorems:

1. If $ f $ is differentiable on $ S $ with $ f ' ( x ) > 0 $ for each $ x \in S $, then $ f $ must be strictly increasing on $ S $;
2. If $ f $ is differentiable on $ S $, then $ f $ is weakly increasing on $ S $ if and only if $ f ' ( x ) \geq 0 $ for each $ x \in S $;
3. If $ f $ is differentiable on $ S $ with $ f ' ( x ) < 0 $ for each $ x \in S $, then $ f $ must be strictly decreasing on $ S $;
4. If $ f $ is differentiable on $ S $, then $ f $ is weakly decreasing on $ S $ if and only if $ f ' ( x ) \leq 0 $ for each $ x \in S $.

(These can all be proved using the Mean-Value Theorem.)

Notice that theorems (1) and (3) are weaker than theorems (2) and (4); they have missing converses that are not true, as you can see from the examples $ x \mapsto x ^ 3 $ and $ x \mapsto - x ^ 3 $.

Now, if you want to define a notion of whether a function $ f $ is strictly/weakly increasing/decreasing at a point $ x $, then you might think, at least if you restrict yourself to differentiable functions, of just looking at $ f ' ( x ) $. Then you'd define increasing/decreasing on a set to mean increasing/decreasing at every point in that set. But because theorems (1) and (3) lack these converses, that would mean changing the definition. I can't prove to you that this is wrong, since definitions vary (such as whether to take the strict or weak version as the default). But the bottom line is that your class seems to be using the same definitions as I am (at least the ‘strictly’ ones).

Answer 4

Let me point out a few things without using the "inc..." word.

I'll assume throughout that $f$ is a real valued function defined on an open subinterval $I$ of the real numbers.

Here is a true statement:

(1) If $f'(x) > 0$ for all $x$ in $I$, then $x<y \implies f(x)<f(y)$ for all $x,y$ in $I$.

This statement is proved in 1st semester calculus by applying the Mean Value Theorem, as I'm sure you know.

The converse of statement (1) is the following false statement:

(2) If $x<y \implies f(x)<f(y)$ for all $x,y$ in $I$, then $f'(x) > 0$ for all $x$ in $I$.

This statement is false because $f(x)=x^3$ is a counterexample.

I'm sure you agree with me so far.

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Now we come to a question of linguistics. The following property of the function $f$ on the interval $I$ occurs in both statements (1) and (2), namely:

$x<y \implies f(x)<f(y)$ for all $x,y$ in $I$.

What terminology shall we adopt to refer to this property in shorthand?

One standard terminology for this property that is adopted by mathematicians is this (and here comes the "inc..." word):

$f$ is increasing on $I$.

There are variants on this. For instance some might insist that we instead say $f$ is strictly increasing on $I$.

And one can certainly have a debate, on a linguistic/philosophical level regarding whether this choice of terminology is appropriate. For example, if one objected to this terminology on the grounds that it is not an accurate description of infinitesmal behavior.... well, perhaps there is a different terminology that you could convince the global mathematical community to adopt (good luck with that).

But what is not up for debate is that (1) is true and (2) is false, because one can prove that (1) is true and (2) is false.

And one thing you can't stop mathematicians from doing is to adopt terminology to refer to mathematical properties in shorthand. Is the terminology always perfect? Do our natural language terms always model exactly all the mathematical subtleties? No. Mathematics and natural language are not always an exact match, so we do the best that we can.

Answer 5

Let me add: a differentiable function $f\colon I\to\mathbb R$ from an interval $I$ to the reals is strictly increasing if and only if $f'(x)\ge0$ for all $x\in I$ and there is no interval $J\subseteq I$ (of positive length) such that $f'(x)=0$ for all $x\in J$ (i.e., the set of points with $f'(x)>0$ is dense in $I$).

Answer 6

In all points other than $x = 0$ $f(x) = x^3$ is clearly increasing because $f’(x) = 3x^2 > 0$

For $x = 0$, you compare $f(x)$ with $f(y)$ for any $y > 0 = x$, and $f(y) > 0 = f(x)$. And you compare with $f(y)$ for any $y < 0 = x$, and $f(y) < 0 = f(x)$. So $f$ is increasing at $x = 0$.

You were misled by $f’(x) = 0$. Once in a while you have a function where this happens because $f(x)$ is constant around $x$. Quite often it happens because f has a maximum or minimum at $x$, so $f$ is increasing up to $x$ and decreasing from $x$ onward. And once in a while you have a function like $x^3$ where $f’(x) = f’’(x) = 0.$