Increasing/Decreasing intervals of a parabola
By Sophia Hammond
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
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$\begingroup$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.
$\endgroup$ 0 $\begingroup$There are two closely-related criteria here:
"The function $f$ is increasing".
"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.
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