Question

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

Explanation:

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

We start by writing the function as \( y = -\frac{1}{2}\sqrt{x + 3} \) where \( x\geq - 3 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap \( x \) and \( y \), so we get \( x=-\frac{1}{2}\sqrt{y + 3} \).

Step3: Solve for \( y \)

First, multiply both sides by - 2: \( - 2x=\sqrt{y + 3} \).
Then, square both sides: \( (-2x)^{2}=y + 3 \), which simplifies to \( 4x^{2}=y + 3 \).
Subtract 3 from both sides: \( y = 4x^{2}-3 \). So the inverse function is \( f^{-1}(x)=4x^{2}-3 \).
Now, we need to determine the domain of the inverse function. The original function \( f(x)=-\frac{1}{2}\sqrt{x + 3} \) 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 \( x\leq\boldsymbol{0} \)