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

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(D) The graph of $f(x)$ is given by the equation $y = (x + 1)^2$ for $x \ge -1$.To find the reflection of the graph of $f(x)$ with respect to the line

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$\sqrt{x} - 1, x \ge 0$

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(D) The graph of $f(x)$ is given by the equation $y = (x + 1)^2$ for $x \ge -1$.To find the reflection of the graph of $f(x)$ with respect to the line $y = x$,we interchange $x$ and $y$.Thus,the equation for the graph of $g(x)$ becomes $x = (y + 1)^2$ for $y \ge -1$.Solving for $y$:$y + 1 = \pm\sqrt{x}$Since $y \ge -1$,we have $y + 1 \ge 0$,so we take the positive root:$y + 1 = \sqrt{x}$$y = \sqrt{x} - 1$Since the domain of $f(x)$ is $x \ge -1$ and its range is $y \ge 0$,the domain of $g(x)$ is $x \ge 0$ and its range is $y \ge -1$.Therefore,$g(x) = \sqrt{x} - 1$ for $x \ge 0$.

Suppose $f(x) = (x + 1)^2$ for $x \ge -1$. If $g(x)$ is the function whose graph is the reflection of the graph of $f(x)$ with respect to the line $y = x$,then $g(x)$ equals

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