Some questions I need help with (1 Viewer)

[Skeptyks]

I might update this thread with questions on top of this one later on that I can't figure out.

1. F(x) = 2x/(x^2+1) -1<=x<=1 (<= is less than equal/greater than equal)

Calculate the total area of the two regions enclosed by y = f(x) and y = f^-1(x)
Now, the next part asks you to evaluate limits 0-1 f^-1(x) which doesn't make sense to me as I can't figure out how to find the area of the parts in the previous question in any other way. I also havn't attempted any inverse trig questions when trying to find f^-1(x) so any explanation on this would be great.

EDIT: 2. Worked out P(2at, at^2), Q(2at/(t^2+1), 2at^2/(t^2+1)) and y/x = t.
Prove that, as P moves on the parabola, Q moves on a circle, and state its centre and radius.

Any help is appreciated, thanks :]

 

[Skeptyks]

bump

 

[Skeptyks]

again

 

[Timske]

F(x) : y = 2x/(x^2+1)
F(x)-1 : x = 2y/(y^2+1) - make y the subject.

 

[Skeptyks]

F(x) : y = 2x/(x^2+1)
F(x)-1 : x = 2y/(y^2+1) - make y the subject.

Embarrassing but, how do I make y the subject... ?

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=x = \frac{2y}{y^2 @plus; 1} \\ x(y^2 @plus; 1) = 2y \\ xy^2 -2y @plus; x = 0 \\ \textup{Use~ Quadratic}\\ y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} = \frac{1 \pm \sqrt{1 -x^2}}{x}\\ y =\frac{1 - \sqrt{1 -x^2}}{x} ~,~ -1\leq x\leq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x = \frac{2y}{y^2 + 1} \\ x(y^2 + 1) = 2y \\ xy^2 -2y + x = 0 \\ \textup{Use~ Quadratic}\\ y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} = \frac{1 \pm \sqrt{1 -x^2}}{x}\\ y =\frac{1 - \sqrt{1 -x^2}}{x} ~,~ -1\leq x\leq 1" title="x = \frac{2y}{y^2 + 1} \\ x(y^2 + 1) = 2y \\ xy^2 -2y + x = 0 \\ \textup{Use~ Quadratic}\\ y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} = \frac{1 \pm \sqrt{1 -x^2}}{x}\\ y =\frac{1 - \sqrt{1 -x^2}}{x} ~,~ -1\leq x\leq 1" /></a>

 

[Skeptyks]

<a href="http://www.codecogs.com/eqnedit.php?latex=x = \frac{2y}{y^2 @plus; 1} \\ x(y^2 @plus; 1) = 2y \\ xy^2 -2y @plus; x = 0 \\ \textup{Use~ Quadratic}\\ y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} = \frac{1 \pm \sqrt{1 -x^2}}{x}\\ y =\frac{1 - \sqrt{1 -x^2}}{x} ~,~ -1\leq x\leq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x = \frac{2y}{y^2 + 1} \\ x(y^2 + 1) = 2y \\ xy^2 -2y + x = 0 \\ \textup{Use~ Quadratic}\\ y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} = \frac{1 \pm \sqrt{1 -x^2}}{x}\\ y =\frac{1 - \sqrt{1 -x^2}}{x} ~,~ -1\leq x\leq 1" title="x = \frac{2y}{y^2 + 1} \\ x(y^2 + 1) = 2y \\ xy^2 -2y + x = 0 \\ \textup{Use~ Quadratic}\\ y = \frac{2 \pm \sqrt{4 - 4x^2}}{2x} = \frac{1 \pm \sqrt{1 -x^2}}{x}\\ y =\frac{1 - \sqrt{1 -x^2}}{x} ~,~ -1\leq x\leq 1" /></a>

Wow, okay. Thanks heaps :]

Does anyone know how to do the first bit of part 1 without evaluating the limits 0-1 of f^-1(x)?