Dear All!

Pleas,e coudl someone help me to get

f(x)=-pi/4 for x<1

and

f(x)=arctan2+arctan3 for x>1

$\displaystyle f(x)=arctan\frac{x+1}{x-1}+arctanx$

Many thanks!!!

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- Jan 16th 2013, 09:43 AMBooarctan
Dear All!

Pleas,e coudl someone help me to get

f(x)=-pi/4 for x<1

and

f(x)=arctan2+arctan3 for x>1

$\displaystyle f(x)=arctan\frac{x+1}{x-1}+arctanx$

Many thanks!!! - Jan 16th 2013, 10:13 AMBariothRe: arctan
you have

$\displaystyle \\ \frac{-\pi}{4}= \arctan{\frac{(x+1)}{(x-1)}} +\arctan(x)\\$

you put Tan everywhere

$\displaystyle \\\tan\frac{-\pi}{4} = \frac{(x+1)}{(x-1)} +(x)\\-1 = \frac{(x+1)}{(x-1)} +(x)\\0 = \frac{(x+1}{x-1} +(x) + 1$

Solve the quadratic formula, you end up with x = 0 and x = (-1) - Jan 16th 2013, 10:59 AMPlatoRe: arctan
As you can I got rid of the <br/> in your post.

That happens if a linefeed in the code. This like this: $\displaystyle

$

Your code should look like

[tex]\\ \frac{-\pi}{4}= \arctan{\frac{(x+1)}{(x-1)}} +\arctan(x)\\[/tex]

you put Tan everywhere

[TEX]\\\tan\frac{-\pi}{4} = \frac{(x+1)}{(x-1)} +(x)\\-1 = \frac{(x+1)}{(x-1)} +(x)\\0 = \frac{(x+1}{x-1} +(x) + 1[/TEX] - Jan 16th 2013, 11:00 AMBariothRe: arctan
Thanks for the info!

- Jan 17th 2013, 08:38 AMBooRe: arctan
Hello!

Many thanks!

But, I should not know in advance that the result is pi/4! I have to calculate it...how?

P.s. Please, can U tell me how did U got rid of arctans putting tan in front of arctan?

Many thanks!!!