Suppose f(x)=(x+1)^{2} for x geq -1. If g(x) | vedclass

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(B) Given that $f(x)=(x+1)^{2}$ for $x \geq -1$. Since $g(x)$ is the reflection of the graph of $f(x)$ in the line $y=x$,$g(x)$ is the inverse functio

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$\sqrt{x}-1$

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(B) Given that $f(x)=(x+1)^{2}$ for $x \geq -1$. Since $g(x)$ is the reflection of the graph of $f(x)$ in the line $y=x$,$g(x)$ is the inverse function of $f(x)$,denoted as $f^{-1}(x)$. To find the inverse,let $y = (x+1)^{2}$. Since $x \geq -1$,we have $x+1 \geq 0$,so taking the square root of both sides gives $\sqrt{y} = x+1$. Solving for $x$,we get $x = \sqrt{y} - 1$. By interchanging $x$ and $y$,we obtain $f^{-1}(x) = \sqrt{x} - 1$. Thus,$g(x) = \sqrt{x} - 1$.

Suppose $f(x)=(x+1)^{2}$ for $x \geq -1$. If $g(x)$ is a function whose graph is the reflection of the graph of $f(x)$ in the line $y=x$,then $g(x) = $

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