Let f :R → R be the functions defined by f (x) = x3 + 5. Then f–1 (x) is

Given that, f(x) = x3 + 5


Let f(x) = y


⇒ y = x3 + 5


⇒ y – 5 = x3


⇒ x3 = y – 5


⇒


⇒ [∵ f(x) = y, ⇒ x = f-1(y)]


⇒

41