What are Direct and Inverse Images in Mathematics?

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Homework Statement



The problem is attached and my attempt for direct image is also attached.
I seek help in inverse image.

Homework Equations





The Attempt at a Solution

 

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You did part (a) correctly!

Part (b):
If h = g \circ f, then h^{-1} = f^{-1} \circ g^{-1}.
Look at the boundaries of G. They will give you the boundaries of its inverse image. This isn't ALWAYS the case, but our function here is pretty simple so it will work out well.
 
Based on your suggestions...
 

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mhmm! and a nice self-assuring way to check that you "almost definitely" got it right is the fact that f(\{x : -2 \le x \le 0\}) = G. I say "almost definitely" because it could happen that you luck out and it works even though your answer's wrong...but let's not worry about that haha. you got it right!
 
It turns out, I am not too dumb, yay! (Thank you for your patience, Raskolnikov.)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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