Let f(x) = (x + 1)^2 - 1 for x ge -1. Then th | vedclass

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(D) Given $f(x) = (x + 1)^2 - 1$ for $x \ge -1$.To find the set $S = \{ x : f(x) = f^{-1}(x) \}$,we note that the solutions to $f(x) = f^{-1}(x)$ are

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$\{0, -1, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2}\}$

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(D) Given $f(x) = (x + 1)^2 - 1$ for $x \ge -1$.To find the set $S = \{ x : f(x) = f^{-1}(x) \}$,we note that the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$ and the points where the graph of $f(x)$ intersects its inverse.Setting $f(x) = x$,we get $(x + 1)^2 - 1 = x$.$x^2 + 2x + 1 - 1 = x \Rightarrow x^2 + x = 0 \Rightarrow x(x + 1) = 0$.Thus,$x = 0$ and $x = -1$ are solutions.However,the equation $f(x) = f^{-1}(x)$ is equivalent to $f(f(x)) = x$ for $x$ in the domain.$( (x + 1)^2 - 1 + 1 )^2 - 1 = x \Rightarrow ((x + 1)^2)^2 - 1 = x$.$(x + 1)^4 - 1 = x \Rightarrow (x + 1)^4 - (x + 1) = 0$.$(x + 1) [ (x + 1)^3 - 1 ] = 0$.This gives $x + 1 = 0 \Rightarrow x = -1$ or $(x + 1)^3 = 1$.The roots of $(x + 1)^3 = 1$ are $x + 1 = 1, \omega, \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$.$x + 1 = 1 \Rightarrow x = 0$.$x + 1 = \frac{-1 + i\sqrt{3}}{2} \Rightarrow x = \frac{-3 + i\sqrt{3}}{2}$.$x + 1 = \frac{-1 - i\sqrt{3}}{2} \Rightarrow x = \frac{-3 - i\sqrt{3}}{2}$.Thus,the set $S = \{ 0, -1, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2} \}$.

Let $f(x) = (x + 1)^2 - 1$ for $x \ge -1$. Then the set $S = \{ x : f(x) = f^{-1}(x) \}$ is

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