Question

3
drag each tile to the correct box. not all tiles will be used.
consider function f.
$f(x) = \sqrt{7x - 21}$
place the steps for finding $f^{-1}(x)$ in the correct order.
$y = \sqrt{7x - 21}$
$\frac{1}{7}(x^2 - 21) = f^{-1}(x)$, where $x \geq 0$
$\frac{1}{7}x^2 + 3 = y$
$\frac{1}{7}x^2 + 3 = f^{-1}(x)$, where $x \geq 0$
$x^2 + 21 = 7y$
$x = \sqrt{7y - 21}$
$x^2 = 7y - 21$
$(x + 21)^2 = 7y$

Explanation:

Step1: Start with original function

We begin by writing the function \( f(x)=\sqrt{7x - 21} \) as \( y=\sqrt{7x - 21} \) (this is the first tile: \( y = \sqrt{7x - 21} \)).

Step2: Swap x and y

To find the inverse, we swap \( x \) and \( y \), so we get \( x=\sqrt{7y - 21} \).

Step3: Square both sides

Square both sides of the equation \( x=\sqrt{7y - 21} \) to eliminate the square root. This gives \( x^{2}=7y - 21 \).

Step4: Solve for y

First, add 21 to both sides: \( x^{2}+21 = 7y \). Then, divide both sides by 7: \( y=\frac{1}{7}x^{2}+3 \).

Step5: Write as inverse function

Replace \( y \) with \( f^{-1}(x) \). Since the original function \( f(x)=\sqrt{7x - 21} \) has a domain \( x\geq3 \) (so the range of \( f(x) \) is \( y\geq0 \)), the domain of \( f^{-1}(x) \) is \( x\geq0 \). So \( f^{-1}(x)=\frac{1}{7}x^{2}+3 \), where \( x\geq0 \).

The correct order of the steps (tiles) is:

1. \( y = \sqrt{7x - 21} \)
2. \( x = \sqrt{7y - 21} \)
3. \( x^{2}=7y - 21 \)
4. \( \frac{1}{7}x^{2}+3 = y \)
5. \( \frac{1}{7}x^{2}+3 = f^{-1}(x) \), where \( x\geq0 \)

(Note: The tiles \( \frac{1}{7}(x^{2}-21)=f^{-1}(x) \), where \( x\geq0 \) and \( (x + 21)^{2}=7y \) are incorrect and not used. The tile \( x^{2}+21 = 7y \) is an intermediate step that can be considered between step 3 and step 4, but the main steps for the inverse function derivation follow the order above.)

Answer:

The correct order of tiles (from first to last step) is:

1. \( y = \sqrt{7x - 21} \)
2. \( x = \sqrt{7y - 21} \)
3. \( x^{2}=7y - 21 \)
4. \( \frac{1}{7}x^{2}+3 = y \)
5. \( \frac{1}{7}x^{2}+3 = f^{-1}(x) \), where \( x\geq0 \)