For the function $f: [1,\infty) \rightarrow [1,\infty)$ defined by $f(x) = (x-1)^4 + 1$, among the two statements:
(I) The set $S = \{x \in [1,\infty) : f(x) = f^{-1}(x)\}$ contains exactly two elements, and
(II) The set $S = \{x \in [1,\infty) : f(x) = f^{-1}(x+1)\}$ is an empty set,

Question

Prepp Team · Accepted Answer

Solution

Statement (I) Analysis

The function is given by $f(x) = (x-1)^4 + 1$ for $x \in [1,\infty)$.

Find the inverse function: Let $y = (x-1)^4 + 1$. Solving for $x$, we get $y-1 = (x-1)^4$. Since $x \ge 1$, $x-1 \ge 0$. Thus, $x-1 = (y-1)^{1/4}$, which yields $x = (y-1)^{1/4} + 1$. So, $f^{-1}(x) = (x-1)^{1/4} + 1$.
Check monotonicity: The derivative is $f'(x) = 4(x-1)^3$. For $x \in [1,\infty)$, $f'(x) \ge 0$, meaning $f(x)$ is strictly increasing.
Solve $f(x) = f^{-1}(x)$: For a strictly increasing function, solutions to $f(x) = f^{-1}(x)$ coincide with solutions to $f(x) = x$. $ (x-1)^4 + 1 = x $ $ (x-1)^4 = x-1 $ Let $v = x-1$. Then $v^4 = v$, or $v^4 - v = 0$. Factoring gives $v(v^3 - 1) = 0$. The solutions are $v=0$ or $v=1$.
Find $x$ values:
- If $v=0$, $x-1=0 \implies x=1$.
- If $v=1$, $x-1=1 \implies x=2$.
Both $x=1$ and $x=2$ belong to the domain $[1,\infty)$.
Conclusion for (I): The set $\{x \in [1,\infty) : f(x) = f^{-1}(x)\}$ contains exactly two elements, $\{1, 2\}$. Statement (I) is TRUE.

Statement (II) Analysis

We analyze the set $S = \{x \in [1,\infty) : f(x) = f^{-1}(x+1)\}$.

Set up the equation: $ f(x) = f^{-1}(x+1) $ $ (x-1)^4 + 1 = ((x+1)-1)^{1/4} + 1 $ $ (x-1)^4 = x^{1/4} $
Substitution: Let $u = x-1$. Since $x \in [1,\infty)$, $u \in [0,\infty)$. Substituting $x = u+1$ into the equation gives: $ u^4 = (u+1)^{1/4} $
Solve for $u$: Raising both sides to the power of 4 yields: $ (u^4)^4 = ((u+1)^{1/4})^4 $ $ u^{16} = u+1 $ Rearranging, we get $u^{16} - u - 1 = 0$. Let $k(u) = u^{16} - u - 1$.
Evaluate $k(u)$ at key points:
- $k(0) = 0^{16} - 0 - 1 = -1$.
- $k(1) = 1^{16} - 1 - 1 = -1$.
- $k(2) = 2^{16} - 2 - 1 = 65536 - 3 = 65533$.
Existence of a solution: Since $k(u)$ is continuous, and $k(1) < 0$ while $k(2) > 0$, the Intermediate Value Theorem guarantees at least one root $u_0$ in the interval $(1, 2)$.
Conclusion for (II): This root $u_0$ corresponds to a solution $x_0 = u_0 + 1$ in the interval $(2, 3)$. Thus, the set $S$ is not empty. Statement (II) is FALSE.