Question

13
drag the tiles to the correct boxes to complete the pairs.
match the one-to-one functions with their inverse functions.
$f(x)=\frac{2x}{3}-17$
$f(x)=x-10$
$f(x)=\sqrt3{2x}$
$f(x)=\frac{x}{5}$
inverse function
function
$f^{-1}(x)=5x$
$f^{-1}(x)=\frac{x^3}{2}$
$f^{-1}(x)=x+10$
$f^{-1}(x)=\frac{3(x+17)}{2}$
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Explanation:

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

Set $y=\frac{x}{5}$, swap $x,y$: $x=\frac{y}{5}$, solve for $y$: $y=5x$

Step2: Find inverse of $f(x)=\sqrt[3]{2x}$

Set $y=\sqrt[3]{2x}$, swap $x,y$: $x=\sqrt[3]{2y}$, cube both sides: $x^3=2y$, solve for $y$: $y=\frac{x^3}{2}$

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

Set $y=x-10$, swap $x,y$: $x=y-10$, solve for $y$: $y=x+10$

Step4: Find inverse of $f(x)=\frac{2x}{3}-17$

Set $y=\frac{2x}{3}-17$, swap $x,y$: $x=\frac{2y}{3}-17$, rearrange: $x+17=\frac{2y}{3}$, solve for $y$: $y=\frac{3(x+17)}{2}$

Answer:

1. $f^{-1}(x)=5x \longrightarrow f(x)=\frac{x}{5}$
2. $f^{-1}(x)=\frac{x^3}{2} \longrightarrow f(x)=\sqrt[3]{2x}$
3. $f^{-1}(x)=x+10 \longrightarrow f(x)=x-10$
4. $f^{-1}(x)=\frac{3(x+17)}{2} \longrightarrow f(x)=\frac{2x}{3}-17$