Is f-1(A ∪ B) the same as f-1(A) ∪ f-1(B)?

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Yes, you are. But I'd feel more comfortable if you tried to prove it rather than taking as an assumption.
 
Have you tried to prove it's true?
 
SammyS said:
Have you tried to prove it's true?

I can prove it using words, not quite there using mathematical symbols, but that part is out of the scope of my class. Thanks guys!
 
You prove two sets are equal by proving that each is a subset of the other. You prove "A" is a subset of "B" by saying "let x\in A", then show "x\in B".

Here, to show that f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B), start by saying "let x\in f^{-1}(A\cup B)". Then y= f(x)\in A\cup B. And that, in turn, means that either y\in A or y \in B. Consider each of those.

Note, by the way, that we are considering the inverse image of sets. None of this implies or requires that f actually have an "inverse".
 

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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