Are g(x) and f(g(x)) both onto?

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if f(y) is one to one, and f(g(x)) is one to one, is g(x) one to one?
if g(x) is one to one, and f(g(x)) is one to one, is f(y) one to one?
if g(x) and f(g(x)) are onto, is f(y) is also onto?
if f(y) and f(g(x)) are onto, is g(x) is also onto?
 
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Well, what do you think? You must show your work before we can help you. Do you know the definitions of one-to-one, and onto functions?
 

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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