1
f(x)=-2tan-1x-pi -∞≤x≤-1
2tan-1x -1≤x≤1
pi-2tan-1x 1≤x≤∞
5 Answers
1y=sin-1(2x/1+x2)
Put x=tanθ
y=sin-1(2tanθ/1+tan2θ)
y=sin-1(sin2θ)
y=2θ
y=2tan-1x
62Honey you have to be a bit careful..
sin-1(sin2θ) is not always 2θ
11gordo... ur graph is perfect.... but can u please explain how u 've defined the curve at the end.... f(x)= pi - 2tan^-1 x when x is between -1 and +1 iis ok... ... but the other conditions....
I'm facing a lot in defining the curve.............
1okay.
for the substitution, x=tan∂
we have ∂=tan-1x
so as x belongs to -∞ to +∞, we will have ∂ from (-pi/2,pi/2)
now we know f(x)=sin-1(sin2∂)
we have
f(x)=sin-1(sin2∂)= 2∂ if -pi/2<2∂<pi/2 or -pi/4<∂<pi/4
= pi-2∂ if pi/2<2∂<pi or pi/4<∂<pi/2
= -2∂-pi if -pi<2∂<-pi/2 or -pi/2<∂<-pi/4
now just convert all ∂ to x substituting ∂=tan-1x
to get f(x) in terms of x
cheers!!