Question

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

Explanation:

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

We start with the function \( f(x)=-\frac{1}{2}\sqrt{x + 3} \), so we write \( y = -\frac{1}{2}\sqrt{x + 3} \).

Step2: Solve for \( x \) in terms of \( y \)

First, multiply both sides by \(-2\) to get \(-2y=\sqrt{x + 3}\).
Then, square both sides: \((-2y)^2 = (\sqrt{x + 3})^2\), which simplifies to \(4y^2=x + 3\).
Next, subtract 3 from both sides: \(x = 4y^2-3\).

Step3: Replace \( x \) with \( f^{-1}(x) \) and \( y \) with \( x \)

We get \( f^{-1}(x)=4x^2 - 3 \).
Now, we need to determine the domain of the inverse function. The original function \( y = -\frac{1}{2}\sqrt{x + 3} \) has a range of \( y\leq0 \) (since the square root is non - negative, multiplied by \(-\frac{1}{2}\) makes it non - positive). So the domain of the inverse function \( f^{-1}(x) \) is \( x\leq0 \).

Answer:

The first box is \( 4 \), the second box is \( 3 \), and the third box is \( 0 \). So \( f^{-1}(x)=\boldsymbol{4}x^{2}-\boldsymbol{3} \), for \( x\leq\boldsymbol{0} \)