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

The correct answer is $\sqrt{x} - 1$

Given that $f(x) = (x+1)^2$ for $x \ge -1$. If $g(x)$ is a function whose graph is a reflection of $f(x)$ in the line $y = x$ then $g(x)$ is inverse of $f(x)$

Let $y = (x+1)^2$

⇒ $\sqrt{y} = (x+1)$

⇒ $x = \sqrt{y} - 1$


Now $f^{-1}(y) = \sqrt{y} - 1$ or

$g(x) = \sqrt{x} - 1$