Increasing/Decreasing intervals of a parabola

I am being told to find the intervals on which the function is increasing or decreasing.

It is a normal positive parabola with the vertex at $(3,0).$ The equation could be $y = (x-3)^2,$ but my confusion comes from the interval on which the parabola is increasing: I would think increasing is $(3,\infty)$ and decreasing is $(-\infty, 3)$.

However the text book teaches to use $[3,\infty)$ and $(-\infty, 3].$ Can you explain this. I thought the function was constant at $x=3$.

Thanks

I like this question, because it touches upon a subtle point.

First, let me say that it would be better to write "the function is flat at $x=3$" than "the function is constant at $x=3$". Any function is constant at any one point, if you look at it that way. Even better than "the function is flat" might be "the function has a horizontal tangent at $x=3$". But anyway.

I agree with the book, when it includes the endpoints of the intervals. Because what is the definition of being increasing on some interval $I$? It is that if $x_1<x_2$ then $f(x_1)<f(x_2)$ for all $x_1,x_2\in I$. With this definition, you see that the function is actually increasing on $[3,\infty)$, and not just on $(3,\infty)$. Indeed, if $x_1=3$, then however I choose $x_2>3$, I will have that $f(x_1)<f(x_2)$. The definition holds true, even if I include the endpoint.

There are two closely-related criteria here:

1. "The function $f$ is increasing".

2. "The function $f$ has positive slope."

In differential calculus, it's easy to get the impression the two are interchangeable. However, they are not, for reasons requiring closer examination.

First, let's agree that $I$ is an interval of real numbers (possibly closed, open, or half-open), and that $f$ is a differentiable real-valued function on $I$. For example, we might have $I = [3, \infty)$ and $f(x) = (x - 3)^{2}$.

Definition 1: The function $f$ is:

• (strictly) increasing on $I$ if, for all numbers $x_{1} < x_{2}$ in $I$, we have $f(x_{1}) < f(x_{2})$.

• non-decreasing on $I$ if, for all $x_{1} < x_{2}$ in $I$, we have $f(x_{1}) \leq f(x_{2})$.

Cautions:

• Some authors use "increasing" to mean "strictly increasing"; others use "increasing" to mean "non-decreasing". Unfortunately, that's not going to change on a time scale shorter than a human lifetime.

• In order to say a function is "increasing" in this sense, the domain must contain at least two points; it makes no sense to say a function is "increasing at a point". (This is close to ordinary English usage: A "trend" requires at least two data points.)

Definition 2: The function $f$ has positive slope on $I$ if for all $x$ in $I$, $f'(x) > 0$.

Theorem: If $f$ has positive slope on an interval $I = (a, b)$, then $f$ is strictly increasing on $[a, b]$.

Proof: It's natural to prove the "contrapositive", that if the conclusion fails, the hypothesis also fails. So, assume $f$ is not strictly increasing on $I$. This means there exist numbers $x_{1} < x_{2}$ in $[a, b]$ such that $f(x_{1}) \geq f(x_{2})$, i.e., $f(x_{2}) - f(x_{1}) \leq 0$. By the Mean Value Theorem, there exists a $z$ with $x_{1} < z < x_{2}$ such that $$f'(z) = \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}} \leq 0.$$ (The inequality holds since the numerator is non-positive and the denominator is strictly positive.) That is, $f$ does not have positive slope on $I = (a, b)$. This completes the proof.

The function $f(x) = (x - 3)^{2}$ satisfies the hypotheses on $[3, b]$ for every $b > 3$, and is therefore strictly increasing on $[3, b]$ for all $b > 3$ even though $f'(3) = 0$.

Generally, it is not true that a strictly increasing differentiable function on an interval has positive slope. The function $f(x) = x^{3}$ is something of a standard counterexample: If $x_{1} < x_{2}$ are real, than $x_{1}^{3} < x_{2}^{3}$; however, $f'(x) = 3x^{2}$, so $f'(0) = 0$. Again, $f$ is strictly increasing on $(-\infty, \infty)$, but $f$ does not have positive slope on this interval. (This example is fairly tame; one can arrange that a strictly increasing function has infinitely many "flat points", even in a bounded interval.)

Fine print: There is a notion of "$f$ is increasing at the point $x_{0}$" if $f'(x_{0}) > 0$. But the information one can glean from this is subtle, not the sort of thing to raise in this context. Particularly, if $f'(x_{0}) > 0$, one cannot deduce there is an interval containing $x_{0}$ on which $f$ is strictly increasing.

• What would the last case look like, where f'(x0)>0 but where f isn't increasing? I struggle to think how we define f'(x0)>0 if there is no other arbitrarily close point which is greater. Commented Jan 22 at 19:05
• @NuclearHoagie The standard example is $f(x)=\frac{x}{2}+x^{2}\sin\frac{1}{x}$ for $x \neq 0$ and $f(0) = 0$. We have $f'(0) = \frac{1}{2}$ from the definition (and a bit of work), but $f'$ is both positive and negative on every neighborhood of $0$, so there is no interval about $0$ on which $f$ is increasing. Commented Jan 26 at 17:45

If you take just the left half of the parabola, you will see that any given point of the graph is lower than all the points that come before it, even for $$(3, 0)$$. This means the point $$(3, 0)$$ should be included in the interval of decrease. Similarly, if you look at just the right half of the parabola, any given point is lower than all the points that come after it, even for $$(3, 0)$$. This means the point $$(3, 0)$$ is also part of the interval of increase.