Few questions (1 Viewer)

rand_althor · Oct 10, 2015

So then would you say f(x) is decreasing for x≤1? Which would mean at x=1, f'(x) is both increasing and decreasing, but not strictly doing either? Or does this just depend on what direction you approach x=1 from (1⁺ vs. 1^-)?

InteGrand · Oct 10, 2015

rand_althor said:
So then would you say f(x) is decreasing for x≤1? Which would mean at x=1, f'(x) is both increasing and decreasing, but not strictly doing either? Or does this just depend on what direction you approach x=1 from (1⁺ vs. 1^-)?

$Yes, $f$ is decreasing on $(-\infty,1]$. Technically these terms (\textsl{increasing} and \textsl{decreasing}) refer to occurring on \emph{intervals}, as seen from the definition I gave earlier. If the interval contains just a single point where the function is defined ($x=1$ in your query, so the interval you are asking about is $[1,1]$), it must be increasing there, as it trivially satisfies the definition of increasing (i.e. for any $x_1,x_2 \in [1,1]$, if $x_2 \geq x_1$, then $f(x_2) \geq f(x_1)$; this is because if $x_1,x_2 \in [1,1]$, then $x_1=1=x_2$, and so $x_2 \geq x_1$, and so $f(x_2) = f(1) \geq f(1) = f(x_1)$, i.e. $f(x_2)\geq f(x_1)$ (as they are equal), and the definition is satisfied). Similarly, it also trivially satisfies the definition of \textsl{decreasing} on any interval that contains only one point. Such an interval consisting of only one point is called a \textsl{degenerate interval}.$

rand_althor · Oct 10, 2015

Ah okay. Thanks for the clarifications!

Few questions (1 Viewer)

InteGrand

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rand_althor

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InteGrand

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rand_althor

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