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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

- Thread starter Chipset3600
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- #1

- Feb 14, 2012

- 79

- Feb 5, 2012

- 1,621

Hi Chipset3600,Hello MHB, the f(0) of this function doesn't exist, so im i wrong or this question dont hv solution?

View attachment 406

Your function seem to have some strange symbols which I don't understand. Is it,

\[f(x)=\begin{cases}\frac{2}{\pi}\mbox{arctan}\left(\frac{x+1}{x^2}\right),&\,x=0\\

(x+1)^{1/\sec(x)}-\frac{\cos 2x}{x+1},&\,x<0\\

\frac{\sec^{2}x}{x.3^x-9x^2},&x>0\\

\end{cases}\]

In that case the definition seem to be problematic since \(f(0)\) is not defined properly. \(f(0)\) should be a constant value whereas in the definition it's not.

Kind Regards,

Sudharaka.

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- #3

- Feb 14, 2012

- 79

and in the exercise is sin^2(x) and not sec^2(x).

Hi Chipset3600,

Your function seem to have some strange symbols which I don't understand. Is it,

\[f(x)=\begin{cases}\frac{2}{\pi}\mbox{arctan}\left(\frac{x+1}{x^2}\right),&\,x=0\\

(x+1)^{1/\sec(x)}-\frac{\cos 2x}{x+1},&\,x<0\\

\frac{\sec^{2}x}{x.3^x-9x^2},&x>0\\

\end{cases}\]

In that case the definition seem to be problematic since \(f(0)\) is not defined properly. \(f(0)\) should be a constant value whereas in the definition it's not.

Kind Regards,

Sudharaka.

- Feb 5, 2012

- 1,621

Good. But the problem here is that the function is not defined at \(x=0\) properly. We know that,Well, my language is portuguese, i forgot the translate of symbols: sen^2(x) = sin^(x), arctg(x)=arctan(x)..

and in the exercise is sin^2(x) and not sec^2(x).

\[\lim_{x\rightarrow 0}\left[\frac{2}{\pi}\mbox{arctan}\left( \frac{x+1}{x^2}\right)\right]=1\]

So maybe the author would have meant ,

\[

f(x)=\begin{cases}1,&\,x=0\\

(x+1)^{1/\sin(x)}-\frac{\cos 2x}{x+1},&\,x<0\\ \frac{\sin^{2}x}{x.3^x-9x^2},&x>0\\

\end{cases}\]