To find the inverse function of \( f(x) = (x - 1)^2 - 4 \) with the domain restricted to \( x \leq 1 \), first, we rewrite the function in terms of \( y \):
\[ y = (x - 1)^2 - 4 \]
Next, solve for \( x \) in terms of \( y \):
- Add 4 to both sides:
\[ y + 4 = (x - 1)^2 \]
- Take the square root of both sides:
\[ \sqrt{y + 4} = |x - 1| \]
Since we have restricted the domain to \( x \leq 1 \), we discard the positive root:
\[ x - 1 = -\sqrt{y + 4} \]
- Solve for \( x \):
\[ x = 1 - \sqrt{y + 4} \]
Now, substituting \( y \) back with \( x \) for the inverse function:
\[ f^{-1}(x) = 1 - \sqrt{x + 4} \]
Now we determine the domain of the inverse. The output of \( f(x) \) when \( x \leq 1 \) can be found by evaluating \( f(1) \):
\[ f(1) = (1 - 1)^2 - 4 = -4 \]
As \( x \) decreases from 1, the function value increases towards positive infinity. Thus, the range of \( f(x) \) (and consequently the domain of \( f^{-1}(x) \)) is:
\[ y \geq -4 \]
Combining this information, the inverse function is:
\[ f^{-1}(x) = 1 - \sqrt{x + 4}, \quad x \geq -4 \]
Thus, the correct response is:
\[ f^{-1}(x) = 1 - \sqrt{x + 4}, \quad x \geq -4 \]