Question

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

• A. $-\sqrt{x} - 1$
• B. $\sqrt{x} + 1$
• C. $\sqrt{x} - 1$
• D. $\sqrt{-x} - 1$

Hint:

Always remember: "Reflection across $y = x$" is the code phrase for "Find the inverse function". The domain restriction ($x \ge 1$) is a critical hint that you must carefully consider whether to take the positive or negative branch when square rooting.

Answer 1

The Correct Option is C

To solve the problem, we need to determine the function \(g(x)\) which is the reflection of \(f(x) = (x + 1)^2\) in the line \(y = x\).

When a function is reflected in the line \(y = x\), the x and y variables are swapped. If \(y = f(x)\), then \(x = g(y)\) is found by solving the equation for x in terms of y.

1. Start with the equation for f(x):

\(y = (x + 1)^2\)

1. Replace x with y and solve for x in terms of y:

\(x = (y + 1)^2\)

1. Reflecting across \(y = x\) gives:

\(y = (x + 1)^2\) is reflected to \(x = (y + 1)^2\)

1. Swap variables and solve for y:
   • \(x = (y + 1)^2\)
   • Solve for x to find \(y\):
   • \(y + 1 = \pm \sqrt{x}\)
   • The function \(f(x)\) is defined for \(x \ge 1\), so \(y \ge 1\) will apply here due to the reflection.
   • When \(x = (y + 1)^2\), the positive square root is appropriate, hence:

\(y + 1 = \sqrt{x}\)

1. Finally, solve for \(y\):
   • \(y = \sqrt{x} - 1\)

This is \(g(x)\), which matches the given correct answer:

\(\sqrt{x} - 1\)

Therefore, \(g(x)\) is correctly given by \(\sqrt{x} - 1\).