Question

The inverse of the function $f(x) = x^2 + 4x + 4, \, x \leq -2$ is $f^{-1}(x) =$

• A. $-2 - \sqrt{x}, \, x \geq 0$
• B. $-2 - \sqrt{x-1}, \, x \geq 1$
• C. $-2 - \sqrt{x}, \, x \geq 2$
• D. $-2 - \sqrt{x}, \, x \geq 4$
• E. $-2 - \sqrt{x}, \, x \geq 5$

Hint:

For inverse of quadratic functions, always use domain restriction to select correct sign of square root.

Answer 1

The Correct Option is A

Concept:
• To find inverse, write $y = f(x)$ and solve for $x$
• Domain restriction helps choose correct branch
• Range of $f(x)$ becomes domain of $f^{-1}(x)$

Step 1: Rewrite function
\[ f(x) = x^2 + 4x + 4 = (x+2)^2 \]

Step 2: Let $y = (x+2)^2$ and solve for $x$
\[ y = (x+2)^2 \Rightarrow x+2 = \pm \sqrt{y} \]

Step 3: Apply domain restriction $x \leq -2$
Since $x \leq -2$, we take negative root: \[ x+2 = -\sqrt{y} \Rightarrow x = -2 - \sqrt{y} \]

Step 4: Find range of $f(x)$
\[ f(x) = (x+2)^2 \geq 0 \] So, range is $y \geq 0$

Step 5: Write inverse function
\[ f^{-1}(x) = -2 - \sqrt{x}, \quad x \geq 0 \] Final Conclusion:
\[ f^{-1}(x) = -2 - \sqrt{x}, \quad x \geq 0 \]