⧼exchistory⧽
22 exercise(s) shown, 0 hidden
BBot
Nov 03'24
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Consider the continuous real-valued function [math]f(x) = x[/math] with domain [math]0 \lt x \lt 1[/math]. Does this function have an absolute maximum point or an absolute minimum point? Why is this function not a counterexample to Theorem?
BBot
Nov 03'24
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[/math]
Let
[[math]]
f(x) = \begin{cases}\frac1x, \textrm{for} \quad \begin{cases}-1 \leq x \lt 0,\\ 0 \lt x \leq 1,\end{cases}\\
0, \mbox{for $x = 0$}.\end{cases}
[[/math]]
This real-valued function is defined on the closed interval [math][-1,1][/math]. Draw the graph of [math]f(x)[/math] and explain why this function has neither absolute maximum nor absolute minimum points.