Question

If \( f(x) = (x+1)^2 - 1, x \ge -1 \), then \( \{x \mid f(x) = f^{-1}(x)\} \) is:

• A. \( \{0, -1\} \)
• B. \( \{-1, 0, 1\} \)
• C. \( \left\{-1, 0, \frac{-3+\sqrt{3}i}{2}, \frac{-3-\sqrt{3}i}{2}\right\} \)
• D. an empty set

Hint:

To find the inverse of a function \( y = f(x) \), swap \( x \) and \( y \) and solve for \( y \). Remember to consider the domain and range of the original function when determining the appropriate branch of the inverse. When \( f(x) = f^{-1}(x) \), the solutions often lie on the line \( y=x \).

Answer 1

The Correct Option is A

Step 1: Find the inverse function \( f^{-1}(x) \).
Given \( f(x) = (x+1)^2 - 1 \) with domain \( x \ge -1 \).
Let \( y = f(x) \). So, \( y = (x+1)^2 - 1 \).
To find the inverse, swap \( x \) and \( y \):
\[ x = (y+1)^2 - 1 \]
Add 1 to both sides:
\[ x+1 = (y+1)^2 \]
Take the square root of both sides.
Since the domain of \( f(x) \) is \( x \ge -1 \), the range of \( f(x) \) is \( f(x) \ge (-1+1)^2 - 1 = 0 - 1 = -1 \).
Therefore, the domain of \( f^{-1}(x) \) is \( x \ge -1 \).
The range of \( f^{-1}(x) \) is the domain of \( f(x) \), which is \( y \ge -1 \).
This means \( y+1 \ge 0 \), so we must take the positive square root:
\[ \sqrt{x+1} = y+1 \]
Subtract 1 from both sides:
\[ y = \sqrt{x+1} - 1 \]
Thus, the inverse function is \( f^{-1}(x) = \sqrt{x+1} - 1 \).
Step 2: Set \( f(x) = f^{-1}(x) \) and solve for \( x \).
\[ (x+1)^2 - 1 = \sqrt{x+1} - 1 \]
Add 1 to both sides:
\[ (x+1)^2 = \sqrt{x+1} \]
Let \( u = \sqrt{x+1} \). Since \( x \ge -1 \), \( x+1 \ge 0 \), so \( u \ge 0 \).
Substitute \( u \) into the equation. Note that \( (x+1)^2 = (\sqrt{x+1})^4 = u^4 \).
\[ u^4 = u \] Rearrange the equation:
\[ u^4 - u = 0 \]
Factor out \( u \):
\[ u(u^3 - 1) = 0 \]
This equation gives two possible cases for \( u \):
\( u = 0 \) \( u^3 - 1 = 0 \implies u^3 = 1 \). Since \( u \) must be a real number (as \( u = \sqrt{x+1} \)), the only real solution is \( u = 1 \). (The other two solutions are complex: \( \frac{-1 \pm i\sqrt{3}}{2} \), but these are not valid for \( u = \sqrt{x+1} \ge 0 \)). 
Step 3: Substitute back \( u = \sqrt{x+1} \) and find \( x \). Case 1: \( u = 0 \)
\[ \sqrt{x+1} = 0 \]
Square both sides:
\[ x+1 = 0 \]
\[ x = -1 \]
Case 2: \( u = 1 \)
\[ \sqrt{x+1} = 1 \]
Square both sides:
\[ x+1 = 1^2 \]
\[ x+1 = 1 \]
\[ x = 0 \]
Step 4: Check if the solutions are valid within the domain.
The domain for \( f(x) \) is \( x \ge -1 \), and the domain for \( f^{-1}(x) \) is also \( x \ge -1 \).
Both solutions, \( x = -1 \) and \( x = 0 \), satisfy the condition \( x \ge -1 \).
Therefore, the set of values of \( x \) for which \( f(x) = f^{-1}(x) \) is \( \{-1, 0\} \).