# Intuition behind a zero derivative

I came across a following statement in my analysis book: A function being strictly increasing doesn't imply that its derivative is always greater than zero. For example, $$(x-2)^5$$ is strictly increasing, while its derivative $$5(x-2)^4$$ is equal to $$0$$ at $$x=2$$.
I'm really having trouble making sense of this. I always thought that a zero derivative at a point $$c$$ means that for points $$x$$ really close to $$c$$, $$f(x) = f(c)$$.
Could someone please explain how this statement makes sense intuitively?

• Draw the (simpler) graph of $~y = x^3,~$ whose derivative is $~= 0 ~$ at the isolated point of $~x = 0.~$ Everywhere else, the derivative is positive. The critical idea is that $~x=0~$ is an isolated point. Commented Nov 19, 2023 at 17:05
• "for x really close to c , f(x) = f(c)" means constant on a small interval around c. What is true is ""for x really close to c , f(x) is really close to f(c)", but that's just continuity. Commented Nov 19, 2023 at 17:09
• The derivative is only zero at a point. For a function to be constant in an interval around that point, all of the derivatives would have to be zero. Commented Nov 19, 2023 at 18:41
• Worse than that: a function can have its derivatives of all orders equal to zero, and still being strictly increasing. The well-known example is $\exp(-1/x^2)$ (for $x \ne 0$, and value $0$ on $0$): it is infinitely differentiable on $0$, and its derivatives of all orders are $0$ on $0$. Commented Nov 19, 2023 at 21:23
• If your function $f(x)$ is continuous at some $x=x_0$ and is increasing when $x<x_0$ and is increasing when $x>x_0$ then it must also be increasing at $x=x_0$ as there is nowhere else it can decrease or stay constant Commented Nov 20, 2023 at 1:26

The "easiest" example to look at is $$f(x)=x^3$$, where the derivative at $$x=0$$ is equal to $$0$$, and indeed the tangent line to the curve is flat.

Simply looking at the graph should help you make intuitive sense of this:

What the derivative being zero tells you is that $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = 0$$ where the expression in the limit is the slope of the line between the points $$(x+h,f(x+h))$$ and $$(x,f(x))$$ on the graph of $$f$$. From the definition of a limit, this means that this slope can be made arbitrarily small. However, that does not imply that it is actually equal to $$0$$ for any non-zero value of $$h$$.

The definition of strictly increasing:

A function $$f(x)$$ is said to be strictly increasing on an interval $$I$$ if $$f(b)>f(a)$$ for all $$b>a$$, where $$a$$,$$b$$ in $$I$$.

This can be rewritten as $$f(b) - f(a) > 0$$ for all $$b>a$$, where $$a$$,$$b$$ in $$I$$.

And the definition of a derivative:

The derivative of a function $$f$$ at $$x$$ is $$\lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$$

The difference between these is that one of them is a statement about the limit of a series, while the other is a statement about the series itself.

We can use a simpler example to drive this home. Consider the sequence $$\frac{1}{n}$$ for $$n > 0$$. This is the sequence $${1, \frac 1 2, \frac 1 3, \frac 1 4, \ldots}$$. The limit of this sequence is 0, but no element in the sequence is 0.

It's the same in your example. If you let $$a=2$$ (and thus $$f(a)=0$$), and try to find some $$b>2$$ such that $$f(b)\le 0$$ by making $$b$$ get closer and closer to 2, you'll see that it can never get there. It will always be just a little bit too large. If we consider the limit of such a sequence, we can easily find one that approaches 0, but the definition of strictly increasing doesn't include a limit.

Let's look at a graph of your function.

This is the function $$(x-2)^5$$. Notice that it is always strictly increasing, as your book says. It never decreases, and it never stays the same over time. That being said, at x values close to 2, it gets pretty close to horizontal. Let's take a closer look.

Look how as x gets closer and closer to 2, the slope (the derivative) gets smaller and smaller. It's not hard to imagine that at the exact point $$x=2$$, the slope is exactly 0, even if only for an instant. And indeed, the value of the derivative $$5(x-2)^4$$ at $$x=2$$ is precisely 0, just as your book claims.

So what can we learn from this? A function doesn't have to have an interval on which it is decreasing, or even an interval on which it is horizontal, to have a zero derivative at some point. The derivative can just barely bump 0 before turning around and increasing once again. As AnalysisStudent says in their excellent answer, the simplest function to exhibit this behavior is $$f(x) = x^3$$. It has a tangent line of $$y=0$$ at $$x=0$$, so the derivative is 0 at that point.

I'm really having trouble making sense of this. I always thought that a zero derivative at a point $$c$$ means that for points $$x$$ really close to $$c$$, $$f(x)=f(c)$$.

When the derivative "just touches" zero at a single point, as in the example, the function has a positive difference over every finite interval. It's "instantaneously" unchanging, but if you look at the derivative, and imagine taking its integral between any two points, you're always going to get some positive values in there, unless those two points are the same point, the one where it's zero. But the integral over a zero span is zero anyway, so the derivative being zero at a single point "doesn't get the chance to matter".

If the derivative was zero over some finite span, or if it ever became negative, then the function would indeed be nonincreasing.

A function being strictly increasing doesn't imply that its derivative is always greater than zero.

This statement is simply trying to emphasize that a zero derivative point doesn't necessarily disqualify a monotonically increasing function from being a strictly increasing function.

The fact of $$\lim_{x\to a}f(x)=f(a)$$ doesn't mean that a point $$b\neq a$$ exists such that $$f(b)=f(a)$$.

The function at point $$x=a$$ has a zero derivative but even that doesn't mean you can show me a point $$b\neq a$$ such that $$f(b)=f(a)$$... Show me "$$b=a+\epsilon$$" and I'll show you that "$$f(b)=f(a+\epsilon)\geq f(a)+\delta>f(a)$$" and thus $$f(b)>f(a)$$.