Let f(x)=(x+1)^2-1, x geq-1. Then {x mid f(x) | vedclass

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(C) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$. To find $f^{-1}(x)$, set $y = (x+1)^2 - 1$. Then $y+1 = (x+1)^2$, so $x+1 = \sqrt{y+1}$ (since $x \geq

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$\{0, -1\}$

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(C) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$. To find $f^{-1}(x)$, set $y = (x+1)^2 - 1$. Then $y+1 = (x+1)^2$, so $x+1 = \sqrt{y+1}$ (since $x \geq -1$), which gives $x = \sqrt{y+1} - 1$. Thus, $f^{-1}(x) = \sqrt{x+1} - 1$. We solve $f(x) = f^{-1}(x)$, which implies $(x+1)^2 - 1 = \sqrt{x+1} - 1$. This simplifies to $(x+1)^2 = \sqrt{x+1}$. Squaring both sides, we get $(x+1)^4 = x+1$, or $(x+1)((x+1)^3 - 1) = 0$. This gives $x+1 = 0$ or $(x+1)^3 = 1$. Case $1$: $x+1 = 0 \Rightarrow x = -1$. Case $2$: $x+1 = 1 \Rightarrow x = 0$. Case $3$: $x+1 = \omega$ or $x+1 = \omega^2$, where $\omega$ is a complex cube root of unity. $x = \omega - 1 = \frac{-1+i\sqrt{3}}{2} - 1 = \frac{-3+i\sqrt{3}}{2}$ and $x = \omega^2 - 1 = \frac{-1-i\sqrt{3}}{2} - 1 = \frac{-3-i\sqrt{3}}{2}$. Since the domain is $x \geq -1$, we only consider real values $x = -1$ and $x = 0$. Thus, the set is $\{0, -1\}$.

Let $f(x)=(x+1)^2-1, x \geq-1$. Then $\{x \mid f(x)=f^{-1}(x)\} =$

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