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!!!
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)
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]