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Intermediate value property for $f(x)= \left\{ \begin{array}{cc} \sin(1/x) & x \neq 0 \\ 0 & x=0 \end{array} \right.$

$g(x)=\frac{1}{x}$ and $h(x)\sin(x)$ are continuous on $]-\infty,0[ \cup]0,\infty[$

$\Rightarrow h\circ g$ is continuous on $]-\infty,0[ \cup]0,\infty[$

$h \circ g $ has IVP on $]-\infty,0[ \cup]0,\infty[$

also $h\circ g (0) = 0$

therefore $h \circ g $ has IVP on $\mathbb{R}$

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  • $\begingroup$ if you mean mean intermediate value theorem certainly not $\endgroup$ – daulomb Dec 8 '17 at 3:54
  • $\begingroup$ This function has intermediate value property right ? $\endgroup$ – So Lo Dec 8 '17 at 4:32
  • $\begingroup$ so it is not possible to apply that theorem. Do you remember the statments of that theorem? $\endgroup$ – daulomb Dec 8 '17 at 4:34
  • $\begingroup$ I am using continuity implies intermediate value property.. how can I show it otherwise ? $\endgroup$ – So Lo Dec 8 '17 at 6:06
  • $\begingroup$ The function is not continuous at zero or on any interval containing zero. To apply IVT you have to a continuous function in some closed and finite interval. Otherwise it doesn't work.. $\endgroup$ – daulomb Dec 8 '17 at 6:37
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Let $f(x)<a<f(y)$ with $x<y$. If $x<0$ and $y>0$ then, since f attains all vales between -1 and 1 on $(o,y)$ [ in fact on $(0,\epsilon)$ for any $\epsilon >0$] it does attain the value $a$ somewhere in $(x,y)$. The remaining cases are all covered by the fact that continuous functions have intermediate value property.

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