Answer:
You know inverse is shifting the positions of x and y
so consider the first function
f(x)= \((x-1)^{5}\) +2
so y = \((x-1)^{5}\)+2
no shift x and y positions
x = \((y-1)^{5}\) +2
Now solve for y which gives
x - 2 = \((y-1)^{5}\)
\(\sqrt[5]{x-2}\) = y - 1
y = \(\sqrt[5]{x-2}\) + 1
Hence it is inverse to do the same for the other function
2) y = \(\sqrt[3]{x}\) -1
shift positions
x = \(\sqrt[3]{y}\) -1
x+1 = \(\sqrt[3]{y}\)
y = \((x+1)^{3}\)
HEnce it is also inverse
