Question

drag the tiles to the correct boxes to complete the pairs.
match the one - to - one functions with their inverse functions.
$f(x)=\frac{2x}{5}-17$ $f(x)=x - 10$ $f(x)=sqrt4{2x}$ $f(x)=\frac{x}{5}$
inverse function function
$f^{-1}(x)=5x$
$f^{-1}(x)=\frac{x^{4}}{2}$
$f^{-1}(x)=x + 10$
$f^{-1}(x)=\frac{5(x + 17)}{2}$
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Explanation:

Step1: Find inverse of \( f(x)=\frac{x}{5} \)

Let \( y = \frac{x}{5} \), swap \( x \) and \( y \): \( x=\frac{y}{5} \), solve for \( y \): \( y = 5x \), so \( f^{-1}(x)=5x \) matches \( f(x)=\frac{x}{5} \).

Step2: Find inverse of \( f(x)=\sqrt[4]{2x} \)

Let \( y=\sqrt[4]{2x} \), swap \( x \) and \( y \): \( x = \sqrt[4]{2y} \), raise to 4th power: \( x^4=2y \), solve for \( y \): \( y=\frac{x^4}{2} \), so \( f^{-1}(x)=\frac{x^4}{2} \) matches \( f(x)=\sqrt[4]{2x} \).

Step3: Find inverse of \( f(x)=x - 10 \)

Let \( y=x - 10 \), swap \( x \) and \( y \): \( x=y - 10 \), solve for \( y \): \( y=x + 10 \), so \( f^{-1}(x)=x + 10 \) matches \( f(x)=x - 10 \).

Step4: Find inverse of \( f(x)=\frac{2x}{5}-17 \)

Let \( y=\frac{2x}{5}-17 \), swap \( x \) and \( y \): \( x=\frac{2y}{5}-17 \), add 17: \( x + 17=\frac{2y}{5} \), multiply by \( \frac{5}{2} \): \( y=\frac{5(x + 17)}{2} \), so \( f^{-1}(x)=\frac{5(x + 17)}{2} \) matches \( f(x)=\frac{2x}{5}-17 \).

Answer:

• \( f^{-1}(x) = 5x \) matches \( f(x)=\frac{x}{5} \)
• \( f^{-1}(x)=\frac{x^4}{2} \) matches \( f(x)=\sqrt[4]{2x} \)
• \( f^{-1}(x)=x + 10 \) matches \( f(x)=x - 10 \)
• \( f^{-1}(x)=\frac{5(x + 17)}{2} \) matches \( f(x)=\frac{2x}{5}-17 \)