To find the inverse of the function \( f(x) = -4\sqrt{x} - 1 \), we'll follow these steps:
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Replace \( f(x) \) with \( y \): \[ y = -4\sqrt{x} - 1 \]
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Solve for \( x \) in terms of \( y \): \[ y + 1 = -4\sqrt{x} \] \[ \sqrt{x} = -\frac{y + 1}{4} \] Now, squaring both sides gives: \[ x = \left(-\frac{y + 1}{4}\right)^2 \] Simplifying further: \[ x = \frac{(y + 1)^2}{16} \]
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Replace \( y \) with \( x \) to express the inverse function: \[ f^{-1}(x) = \frac{(x + 1)^2}{16} \]
Next, we need to find the appropriate restriction on \( x \):
- The original function \( f(x) \) has a domain of \( [0, \infty) \) because \( \sqrt{x} \) is defined for \( x \geq 0 \).
- The corresponding range of \( f(x) \) goes from \(-1\) downwards (as \( x \) increases, \( f(x) \) decreases).
Thus, the range of \( f(x) \) is \( (-\infty, -1] \), which means the inverse function should be valid for \( x \leq -1 \).
Therefore, the inverse function is:
\[ f^{-1}(x) = \frac{(x + 1)^2}{16}, \quad x \leq -1 \]
So the correct response is:
\[ \boxed{f^{-1}(x) = \frac{(x + 1)^2}{16}, \quad x \leq -1} \]