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Which of the following are true and which false? Give proofs or counterexamples.

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i). If $\displaystyle x \rightarrow f(x)^2$ (this is a mapping, I couldn't quite find the right arrow) is continous at c then f is differentiable at c.

I said this was false and my counterexample was $\displaystyle f(x)=| \sqrt{x}|$.

This function is continous at 0 but $\displaystyle f(x)^2=|x|$ which is not differentiable at 0.

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ii). If $\displaystyle \exists \ g:(a,b) \rightarrow \mathbb{R}$ and $\displaystyle Z:(a,b) \rightarrow \mathbb{R}$ such that $\displaystyle f(x)=g(x)Z(x)$ then f is differentiable at c if:

a). Z is bounded

b). g is differentiable at c with g'(c)=0 and

c). g(c)=0;

I'm not really sure about this one.

I think it's true but I don't know how to prove it. (Wondering)

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iii). In ii) f is differentiable assuming only (a) and (c).

I put false and used the counterexample $\displaystyle f(x)=x \sin \left( \frac{1}{x} \right)$ when c=0.

$\displaystyle Z(x)= \sin \left( \frac{1}{x} \right)$ and $\displaystyle g(x)=x$.

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In ii). f is differentiable assuming only (a) and (b).

Once again I put false and used the counterexample $\displaystyle f(x)=5 \sin \left( \frac{1}{x} \right)$ where c=0.

Where $\displaystyle g(x)=5$ and $\displaystyle Z(x)= \sin \left( \frac{1}{x} \right)$

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In (ii). f is differentiable assuming only (b) and (c).

This is false because $\displaystyle Z(x)$ could be discontinous.

My counterexample was $\displaystyle f(x)=x^2. \begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\

0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}$ at c=0.

Here $\displaystyle g(x)=x^2$ and $\displaystyle Z(x)=\begin{cases} -1 \ \ \ \ \ x \in{\mathbb{Q}}\\

0 \ \ \ \ \ x \in \mathbb{R}-{\mathbb{Q}} \end{cases}$.

are these correct?