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Let $f:(0,1)\to (0,1)$ be a continuously differentiable function.Then we can conclude that

  • $g=\frac{1}{f}$ is a continuous function on $(0,1)$.
  • $g=\frac{1}{f}$ is a continuously differentiable function on $(0,1)$.
  • $g=\frac{1}{f}$ is a uniformly continuous function on $(0,1)$.
  • $h$ defined by $h(x)=x(1-x)f(x);x\in (0,1)$ is uniformly continuous.

My effort: Since $f(x)\neq 0\forall x$ so $\frac{1}{f}$is a continuously differentiable function. Hence 1 and 2 are correct.

For 3 I have chosen $f(x)=x$

For 4 ;We know that $h$ will be uniformly continuous over $(0,1)$ iff it can be extended continuously over $[0,1]$.Moreover $h$ is a differentiable function so if I can show that $h^{'}$ i.e. if I can show that $f^{'}$is bounded then it will be uniformly continuous.

But how can I conclude that $f^{'}$ is bounded?Please help me out.

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    $\begingroup$ The first three of your questions are identical. For the last, look at functions like $\frac 1{x^2}$. $\endgroup$ – lulu Aug 8 '16 at 11:55
  • $\begingroup$ I have done required edits;your example does not work $\endgroup$ – Learnmore Aug 8 '16 at 12:01
  • $\begingroup$ Quite right as to bad example. Ignored condition on range. $\endgroup$ – lulu Aug 8 '16 at 12:24
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Since $f$ is bounded, the factor $x(1-x)$ forces the continuity in $x=0$ and $x=1$. A continuous function on a compact set is automatically uniformly continuous.

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  • $\begingroup$ Thank you very much for the details $\endgroup$ – Learnmore Aug 8 '16 at 12:41

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