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This was brought up by another student in one of my pre-calculus classes.

The graph was a simple quadratic $x^2$. The teacher stated that the graph was decreasing from $(-\infty,0)$, and increasing from $(0, \infty)$.

Why would zero not be included? i.e: decr. $(-\infty,0]$ and incr. $[0, \infty)$

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Because for $f(x)$ to be decreasing $f'(x)<0$ And for increasing $f'(x)>0$ But at $x=0$, $f'(x)=0$ hence it's neither decreasing nor increasing at $x=0$.

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Generally the $0$ is not included because the function is not decreasing (or increasing) at $0$.

It would be accurate, however to say that $y=x^2$ is non-increasing on the interval $(-\infty,0]$.

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  • $\begingroup$ But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it? $\endgroup$ – Anonymous Sep 21 '16 at 0:08
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    $\begingroup$ The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-\infty,0]$. Another definition says that $f$ decreases on $I$ if for $x\in I$, $f^\prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-\infty,0]$.So the 'correct' answer depends upon which definition is being used. $\endgroup$ – John Wayland Bales Sep 21 '16 at 21:48

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