Question

type the correct answer in each box. use numerals instead of words.
what is the inverse of this function?
$f(x)=\frac{-1}{2}\sqrt{x + 3}, x \geq -3$
$f^{-1}(x) = \quad x^2 - \quad$, for $x \leq$

Explanation:

Step1: Replace $f(x)$ with $y$

$y = -\frac{1}{2}\sqrt{x + 3}$

Step2: Swap $x$ and $y$

$x = -\frac{1}{2}\sqrt{y + 3}$

Step3: Isolate the square root

Multiply both sides by $-2$:
$-2x = \sqrt{y + 3}$

Step4: Eliminate the square root

Square both sides:
$(-2x)^2 = (\sqrt{y + 3})^2$
$4x^2 = y + 3$

Step5: Solve for $y$

Subtract 3 from both sides:
$y = 4x^2 - 3$

Step6: Find domain of inverse

Original function $f(x)$ has range $y \leq 0$, so inverse domain is $x \leq 0$.

Answer:

$f^{-1}(x) = 4x^2 - 3$, for $x \leq 0$
The blanks are filled with 4, 3, and 0 respectively.