$\displaystyle g(x) = \sqrt{1 - x^2}$ I know you set it into a rational equality so. . . $\displaystyle 1 - x^2 > 0$ And now what?
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Originally Posted by Macleef $\displaystyle 1 - x^2 > 0$ And now what? $\displaystyle 1 > x^2 $?
Following Plato's Hint we have $\displaystyle |x|\le1\implies -1\le x\le1.$ (It's actually greater or equal than zero, the condition of course.)
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