0
$\begingroup$

In my school text book, It say that the quadratic function is increasing and decreasing at some intervals based on the function, however both intervals didn't include the vertex and it was't discussed at all.

Should the vertex be constant because its slop is equal to zero? Or should I just leave it and not include it in any interval like the text book proposed? And how come I can leave it while it is from the domain of the function?

Example:

Discuss the monotonicity of the function $f(x)= {x}^{2}$:

$f(x) \text{ increases at }x \in ]0,\infty[$

$f(x) \text{ decreases at }x \in ]-\infty,0[$

As you can see they didn't include zero in any of the intervals.

$\endgroup$
1
$\begingroup$

Should the vertex be constant because its slop is equal to zero?

Monotonicity is generally defined on a subset of the domain which includes multiple points e.g. an interval, so (again, in general) it doesn't make sense to refer to monotonicity at one single point.

(As a side note, it does make sense for differentiable functions to associate the derivative at a point with the "rate of change" in some sense, and it's common to say that $f(x)$ is increasing at $x_0$ if $f'(x) \gt 0$ however that's more of a casual language license, and also it's a sufficient condition but not a necessary one, for example $f(x) = x^3$ is strictly increasing on $\mathbb{R}$ but $f'(0)=0$.)

Back to the question, what the book says is correct. However, it is equally correct, and in fact a stronger statement, to say that $f(x)$ is strictly decreasing on $(-\infty,0\,]$ and strictly increasing on $[\,0,\infty)$ where both intervals include $0$.

And how come I can leave it while it is from the domain of the function?

There is no requirement (or guarantee) that the intervals of monotonicity cover the entire domain.

For example $f(x) = x \cdot sin \frac{1}{x}$ for $x \ne 0$ with $f(0)=0$ is a continuous function which is not monotonic on any interval that includes $0$. For an example of a discontinuous function which is not monotonic on any interval consider the indicator function of $\mathbb{Q}$.

$\endgroup$
  • $\begingroup$ Thanks, do you mind telling me why you didn't use a square bracket on the infinity and negative infinity $(-\infty,0\,]$? $\endgroup$ – Omar Ahmad Nov 13 '16 at 17:27
  • 1
    $\begingroup$ Infinities are not real numbers ($\pm \infty \not \in \mathbb{R}$) so they can't belong to a real interval. The notation $x \in (-\infty,0]$ simply means $x \le 0$. $\endgroup$ – dxiv Nov 13 '16 at 17:30
1
$\begingroup$

If you would choose to put zero in one of the two intervals, which one would you choose?

At the vertex the slope is zero, so the function is neither increasing nor decreasing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.