# True or false analysis.

• May 9th 2009, 03:52 AM
Showcase_22
True or false analysis.
Quote:

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

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.

Quote:

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

Quote:

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 $\displaystyle g(x)=x$.

Quote:

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

Quote:

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?
• May 9th 2009, 08:41 AM
running-gag
Quote:

Originally Posted by Showcase_22
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.

You mean $\displaystyle f(x)=\sqrt{|x|}$ I suppose ?
If so it is OK
• May 9th 2009, 08:44 AM
running-gag
Quote:

Originally Posted by Showcase_22

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

Yes it is true

$\displaystyle \frac{f(x) - f(c)}{x-c} = \frac{g(x)Z(x) - g(c)Z(c)}{x-c} = \frac{g(x)Z(x)}{x-c} = Z(x) \:\frac{g(x)}{x-c} = Z(x) \:\frac{g(x) - g(c)}{x-c}$

I think that you can finish.
This will give you ideas for the other cases.
• May 9th 2009, 08:53 AM
running-gag
Quote:

Originally Posted by Showcase_22
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 $\displaystyle g(x)=x$.

I do not agree with your counter-example since g is not defined in 0 (even if its limit is 0).

Take $\displaystyle Z(x) = 1 \mbox{ and } g(x) = \sqrt{|x|}$ over (-1,1) for instance

Z is bounded and g(0)=0 therefore (a) and (c) are verified

$\displaystyle f(x) = \sqrt{|x|}$ is not differentiable in 0