Finding Inverses of F(x) on Restricted Intervals

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Homework Statement


Let F(x)=x + 1 / x

a.) sketch graph
b.)From the Graph, it appears that f(x) becomes 1:1 when we restrict its domain to each of the four intervals (-infinity, -1], [-1,0), (0,1] and [1, infinity). For Each of these intervals, find a formula for the inverse of f(x) restricted to the interval.


Homework Equations



Well i sketched the graph which is 2 asymptotes in the X and Y Direction in quadrant 1 and 3... but where do i go from here? i really don't understand the question... just need some help as to where i start and where i end... thank you.



The Attempt at a Solution

 
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The question is, can you express x in terms of y by solving y = F(x), for each of the intervals?
 
Yes You can solve x in terms of y...sorry i don't know what to do next
 
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Solve x in terms of y. I will represent the solution as x = g(y), where g(y) is for you to find. Then see over which intervals (on the x axis) g is a well-defined function. (That is, make sure that g is defined over each of the four subintervals.)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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