$If f (x) = x^{5} + x^{3} + x, find f^{- 1} (3) and f (f^{- 1} (5)) .$ ext{If } f(x) = x^5 + x^3 + x, ext{ find } f^{-1}(3) ext{ and } f\left(f^{-1}(5) ight). (a) $f^{- 1} (3) =$ f^{-1}(3)=[Fill in the blank], (b) $f (f^{- 1} (5)) =$ f\left(f^{-1}(5) ight)=[Fill in the blank],

Question

If f (x) = x^{5} + x^{3} + x, find f^{- 1} (3) and f (f^{- 1} (5)) .

ext{If } f(x) = x^5 + x^3 + x, ext{ find } f^{-1}(3) ext{ and } f\left(f^{-1}(5) ight). (a)

f^{- 1} (3) =

f^{-1}(3)=[Fill in the blank], (b)

f (f^{- 1} (5)) =

f\left(f^{-1}(5) ight)=[Fill in the blank],

BlackTom AI · Accepted Answer

Let's walk through the problem step by step and keep the logic explicit for each part.

- First, recall the function given: f(x) = x^5 + x^3 + x.
  This is a polynomial with odd degree and positive leading coefficient, and its derivative f'(x) = 5x^4 + 3x^2 + 1 is always positive for all real x. Therefore, f is strictly increasing on the entire real line and is one-to-one, so an inverse function f^{-1} exists and is defined for every real y in the range of f (which is all real numbers since a strictly increasing polynomial with odd degree maps R onto R).

- Part (a): determine f^{-1}(3).
  Since f is invertible, f^{-1}(3) is the unique real number x such that f(x) = 3.
  Check x = 1: f(1) = 1^5 + 1^3 + 1 = 1 + 1 + 1 = 3, which matches. Because f is strictly increasing, this x = 1 is the only solution, so f^{-1}(3) = 1.

- Part (b): determine f(f^{-1}(5)).
  Let y = 5. Since f is invertible, by the basic inverse function property we have f(f^{-1}(y)) = y for every y in the domain of f^{-1} (i.e., every y in the range of f).
  In particular, 5 is in the range of f (as f is onto R due to being strictly increasing and unbounded in both directions), so f(f^{-1}(5)) = 5.
  An intuitive way to see this is: if a = f^{-1}(5), then f(a) = 5 by definition, and applying f again to that input gives f(a) = 5.

- Summarizing the reasoning in plain terms:
  • f is one-to-one, so its inverse exists and yields unique values for f^{-1}(3) and f^{-1}(5).
  • The value x = 1 satisfies f(x) = 3, hence f^{-1}(3) = 1.
  • The composition f(f^{-1}(5)) must return 5 because the inverse recovers the original input value under the forward function.

- Note on common misconceptions:
  Some students worry about the inverse of a non-linear function, but the key here is the monotonicity (strictly increasing) ensuring invertibility. Others might doubt the second part, but the identity f(f^{-1}(y)) = y holds for any y in the image of f, so as long as 5 lies in that image, the result is exactly 5.

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