What is the inverse of h(y) where y=|x|

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I'm preparing for my Statistics and Probability exam tomorrow, and I have a quick question:

What is the inverse of h(y) where y=|x|. (just to make sure, h'(x)=1, right?)
 
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h'(x) is not 1 for all x. Draw the graph of |x| and think again. Also do you know the horizontal line test for whether a function is invertible?
 
Well when I draw the graph for |x| i get like a graph like this starting at the origin \|/ , and I'm not sure how to find the inverse or by using the horizontal line test? is that like one-to-one function type?
 
This is the problem that I'm doing:
Suppose that Z is a standard normal random variable: i.e. Z~N(0,1).
a) Find the distribution of X=|Z| .
b) What is the density of X?
c) Find the distribution of Y=X^2
d) What is the joint distribution of X and Y?
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For a) P(X</=x) = P(|Z|</= x) = P(-z</= x </= z) = P(x</=z) - P(x</= -z)
I'm stuck here...

For b) is the density function for this the same as the one that is given as the definition. I mean fx(x)=[1/root2(pie)]e^(-x^2)/2?

For c) I got N(0, 2root(y)) as the distribution of Y=X^2.
 
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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