If f(x) = (x+1)^2 - 1 for x geq -1,then find | vedclass

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

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$\{0, -1\}$

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(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$. To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Then $y+1 = (x+1)^2$. Since $x \geq -1$,$x+1 = \sqrt{y+1}$,so $x = \sqrt{y+1} - 1$. Thus,$f^{-1}(x) = \sqrt{x+1} - 1$. The equation $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ because $f$ is an increasing function for $x \geq -1$. So,$(x+1)^2 - 1 = x$. $x^2 + 2x + 1 - 1 = x$. $x^2 + x = 0$. $x(x+1) = 0$. This gives $x = 0$ or $x = -1$. Both values satisfy the condition $x \geq -1$. Therefore,the set is $\{0, -1\}$.

If $f(x) = (x+1)^2 - 1$ for $x \geq -1$,then find the set $\{x \mid f(x) = f^{-1}(x)\}$.

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