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Answer 1 · 2017-03-24T17:27:04

The inverse is $= \sqrt[5]{x - 2}$ $=root(5)(x-2)$

Let $y = x^{5} + 2$ $y=x^5+2$

Then,

$x^{5} = y - 2$ $x^5=y-2$

$x = \sqrt[5]{y - 2}$ $x=root(5)(y-2)$

Therefore,

$f^{- 1} (x) = \sqrt[5]{x - 2}$ $f^-1(x)=root(5)(x-2)$

Verification

$f (f^{- 1} (x)) = f (\sqrt[5]{x - 2}) = {(\sqrt[5]{x - 2})}^{5} + 2$ $f(f^-1(x))=f(root(5)(x-2))=(root(5)(x-2))^5+2$

$= x - 2 + 2 = x$ $=x-2+2=x$