Why is union sign not allowed to denote intervals of function increasing?

Question

On the calculus exam, we were given a function $$f(x) = \frac{2x^2+x-1}{x^2-1} $$ and asked to find intervals of function increasing and decreasing. In the post-exam solutions PDF, it was written that f(x) is increasing on $$(-\infty; -1), (-1; 1), (1; +\infty)$$ And specifically noted that writing the function increase interval as $$(-\infty; -1) \cup (-1; 1) \cup (1; +\infty)$$ is wrong and unacceptable, however, I fail to understand why so.

So, why is union-sign notation incorrect?

Answer 1

The function is decreasing (not increasing, as you said) on each of those three intervals separately. But it is not decreasing on their union because, for example, $f(5)>f(0)$.

Consider a similar case: my bank balance is decreasing from December 9 through December 14, and from December 16 through December 21. Is it therefore decreasing on the entire interval? No, because on December 15 I received my salary, so my bank balance increased sharply between December 14 and 16.

Answer 2

Because $(-\infty,-1)\cup(-1,1)\cup(1,+\infty)$ is not an interval.

Answer 3

I have never seen one write an interval like $(a;b)$. The standard notation is $(a,b)$.

Consider a simpler example: $g(x)=\frac{-1}{x}$. Note that $g$ is increasing on the interval $(-\infty,0)$ and also increasing on the interval $(0,\infty)$. However, $g$ is not increasing on the set $$ S=(-\infty,0)\cup (0,\infty). $$

Answer 4

Introduction

So, we are considering the function $f$ defined by

$$f(x) = \frac{2x^2+x-1}{x^2-1}.$$

Clearly, this function is defined everywhere except at $x = \pm 1$, so its domain is $D_f = \mathbf{R} \setminus \{-1, +1\}.$

Let us draw its graph, just so we can see very clearly what is going on:

Now, we are interested in knowing where the function is decreasing. Obviously, to answer this question, we need to know exactly what the word "decreasing" means in this context.

What does the word mean?

Likely your calculus textbook contains a definition very similar to this one:

A function $f : D_f \to \mathbf{R}$ is decreasing on the set $U \subset D_f$ if, for all $x, y \in U$, it holds that $x < y \Rightarrow f(x) \ge f(y)$.

So, fundamentally, the concept of "decreasing" applies not to a function alone -- or to a function at a particular point --, but to a function and a subset of its domain ($U$ in the definition above). For instance, it is meaningless to say that "$f$ is decreasing at $x = 5$" [you do not have two points to compare¹!], but it makes sense to say that "$f$ is decreasing on $(-3, -2)$". (And that would be true in our case.)

Let's apply the definition

Now it is easy to see that our function $f$ is indeed decreasing on each of the subsets $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. For example, if you pick any two numbers $x$ and $y$ in $(-1, 1)$, it is obvious that if $x < y$, then $f(x) \ge f(y)$.

Notice how beautifully the precise definition captures the essence of our intuitive idea of what it means to be "decreasing"!²

But with this precise knowledge, it also becomes very easy for us to see why $f$ isn't decreasing on $(-\infty, -1) \cup (-1, 1) \cup (1, \infty) = \mathbf{R} \setminus \{-1, +1\} = D_f$.

Think about it for a while.

For example, take $x = -2$ and $y = 2$. Clearly $x < y$, but $f(x) \ge f(y)$ doesn't hold. [Intuitively: when we moved to the right, the value of the function increased!]

So it is very much not the case that $f$ is decreasing on $D_f$.

A small note on terminology

If the exam question asked for "the intervals where the function is decreasing", you should probably answer it by giving one or more intervals -- and not sets that aren't intervals. But the union clearly isn't an interval, so this also makes it obvious that that answer is incorrect.

[For example, there are no two real numbers $a$ and $b$ such that

$$\mathbf{R} \setminus \{-1, +1\} = (a, b)$$

so the union clearly isn't a bounded open interval.]

A subtle difference

A different exam question could have asked you to find the points in $D_f$ where the function's derivative is non-positive. This is a related but different question. The derivative of a function is also a function, and it has a value at each point where it is defined. So in contrast to the discussion above, it does make sense to say that the derivative of $f$ is non-positive at the particular point $x = 5$, say.

If you compute the derivative of $f$, you find a function $f'$ that is also defined on $D_f$. And its value is non-positive (even negative) everywhere. So it is true that the derivative is non-positive in each of the subsets $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. But this immediately implies that the derivative is non-positive in their union $D_f$.

Consequently, this hypothetical exam question could (indeed, should) be answered by giving the union in its most simplified form.

Footnotes

¹ Exercise (very technical): What do you have to say about the statement "$f$ is decreasing on $\{5\}$"?

² Yeah, this is one of the reasons why I love mathematics. And the $\epsilon\delta$ definitions in calculus ... Beautiful.

Answer 5

The existing answers have focused on the fact that $f$ is not increasing from one interval to the next. An even better example would be $g(x)$ defined as $-1/x$ when $x<0$ and $\sqrt{x}$ when $x\ge 0$. Then $g$ is increasing on $(-\infty,0);[0,\infty)$. But we definitely can't say that $g$ is increasing on $(-\infty,0)\cup[0,\infty)=(-\infty,\infty)$.