drag the tiles to the correct boxes to complete the pairs. match the one - to - one functions with their…

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 $f^{-1}(x)=5x$ $f^{-1}(x)=\frac{x^{3}}{2}$ $f^{-1}(x)=x + 10$ $f^{-1}(x)=\frac{3(x + 17)}{2}$

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 $f^{-1}(x)=5x$ $f^{-1}(x)=\frac{x^{3}}{2}$ $f^{-1}(x)=x + 10$ $f^{-1}(x)=\frac{3(x + 17)}{2}$

Answer

Explanation:

Step1: Find inverse of $f(x)=\frac{2x}{3}-17$

Let $y = \frac{2x}{3}-17$. Solve for $x$: $y + 17=\frac{2x}{3}$, then $x=\frac{3(y + 17)}{2}$. So $f^{-1}(x)=\frac{3(x + 17)}{2}$.

Step2: Find inverse of $f(x)=x - 10$

Let $y=x - 10$. Solve for $x$: $x=y + 10$. So $f^{-1}(x)=x + 10$.

Step3: Find inverse of $f(x)=\sqrt[3]{2x}$

Let $y=\sqrt[3]{2x}$. Cube both sides: $y^{3}=2x$, then $x=\frac{y^{3}}{2}$. So $f^{-1}(x)=\frac{x^{3}}{2}$.

Step4: Find inverse of $f(x)=\frac{x}{5}$

Let $y=\frac{x}{5}$. Solve for $x$: $x = 5y$. So $f^{-1}(x)=5x$.

Answer:

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