As given by the question
There are given that the function:
[tex]f(x)=\sqrt[16]{x}[/tex]
Now,
To find the inverse, first interchange the function f(x) to y:
So,
[tex]\begin{gathered} f(x)=\sqrt[16]{x} \\ y=\sqrt[16]{x} \end{gathered}[/tex]
Then,
Interchange the variable:
So,
[tex]\begin{gathered} y=\sqrt[16]{x} \\ x=\sqrt[16]{y} \end{gathered}[/tex]
Then,
Solve the above equation for y
So,
[tex]\begin{gathered} x=\sqrt[16y]{\square} \\ x=(y)^{\frac{1}{16}} \\ (x)^{16}=(y)^{\frac{1}{16}\times16} \\ y=x^{16} \end{gathered}[/tex]
Then,
The inverse function is :
[tex]f^{-1}(x)=x^{16}[/tex]
Hence, the correct option is (A) and the value is shown below:
[tex]f^{-1}(x)=x^{16}[/tex]
(B):
For verify, put inverse function into the given function:
So,
[tex]\begin{gathered} f(x)=\sqrt[16]{x} \\ f(f^{-1}(x))=\sqrt[16]{(x^{16})} \\ f(f^{-1}(x))=(x^{16})^{\frac{1}{16}} \\ f(f^{-1}(x))=x \end{gathered}[/tex]
And,
Put the value of f(x) into the inverse function:
So,
[tex]\begin{gathered} f^{-1}(x)=x^{16} \\ f^{-1}(f(x))=(\sqrt[16]{x})^{16} \\ f^{-1}(f(x))=(x^{\frac{1}{16}})^{16} \\ f^{-1}(f(x))=x \end{gathered}[/tex]
Hence proved.
