Answer
(a)
[tex]f^{-1}(x)=\frac{6}{x}[/tex]
Explanation
Given function:
[tex]f(x)=\frac{6}{x}[/tex]
(a) To find f⁻¹(x)
[tex]\begin{gathered} \text{Let y }=f(x) \\ \text{This implies, y }=\frac{6}{x} \\ x=\frac{6}{y} \\ \text{Note that x }=f^{-1}(y) \\ f^{-1}(y)=\frac{6}{y} \\ \therefore f^{-1}(x)=\frac{6}{x} \end{gathered}[/tex]
(b) To show that
[tex]f(f^{-1}(x))=x\text{ and }f^{-1}(f(x))=x[/tex][tex]\begin{gathered} To\text{ find }f(f^{-1}(x)) \\ \text{Subtitute }x=f^{-1}(x)\text{ into }f(f^{-1}(x)) \\ \text{Since }f(x)=\frac{6}{x},\text{ it follows that} \\ f(f^{-1}(x))=\frac{6}{f^{-1}(x)}=6\div\frac{6}{x}=6\times\frac{x}{6}=x \end{gathered}[/tex]
Also,
[tex]\begin{gathered} to\text{ find }f^{-1}(f(x)) \\ \text{Substitute }x=^{}f(x)\text{ into }f^{-1}(f(x)) \\ \text{Since }f^{-1}(x)=\frac{6}{x},\text{ it follows that} \\ f^{-1}(f(x))=\frac{6}{f(x)}=6\div\frac{6}{x}=6\times\frac{x}{6}=x \end{gathered}[/tex]
Therefore, it is verified that
[tex]f(f^{-1}(x))=f^{-1}(f(x))=x[/tex]
