Question:

Step 1: Analyze $f(x)$ on interval $(1,2)$ For $x \in (1,2)$, we have: $[x-1] = 0$ (since $0 < x-1 < 1$) $[x] = 1$ $\{x\} = x-1$ Substitute these into the definition of $f(x) = [x-1] + \{x\}^{[x]}$: $$f(x) = 0 + (x-1)^1 = x-1$$ The range of $f(x)$ for $x \in (1,2)$ is $(0,1)$. Step 2: Find $f^{-1}(x)$ for $x \in (0,1)$ Let $y = f(x)$. From Step 1, we have $y = x-1$. Solving for $x$ in terms of $y$: $$x = y+1$$ Thus, the inverse function is $f^{-1}(y) = y+1$. Replacing $y$ with $x$, we obtain: $$f^{-1}(x) = x+1 \quad \text{for } x \in (0,1)$$ Step 3: Analyze $f(x)$ on interval $(2,3)$ For $x \in (2,3)$, we have: $[x-1] = 1$ (since $1 < x-1 < 2$) $[x] = 2$ $\{x\} = x-2$ Substitute these into the definition of $f(x) = [x-1] + \{x\}^{[x]}$: $$f(x) = 1 + (x-2)^2$$ To find the range of $f(x)$ for $x \in (2,3)$: As $x \to 2^+$, $f(x) \to 1 + (2-2)^2 = 1$. As $x \to 3^-$, $f(x) \to 1 + (3-2)^2 = 1+1 = 2$. So the range of $f(x)$ for $x \in (2,3)$ is $[1,2)$. Step 4: Find $f^{-1}(x)$ for $x \in [1,2)$ Let $y = f(x)$. From Step 3, we have $y = 1 + (x-2)^2$. Rearranging the equation: $$y-1 = (x-2)^2$$ Since $x \in (2,3)$, $x-2$ is positive. Therefore, we take the positive square root: $$x-2 = \sqrt{y-1}$$ Solving for $x$ in terms of $y$: $$x = 2 + \sqrt{y-1}$$ Thus, the inverse function is $f^{-1}(y) = 2 + \sqrt{y-1}$. Replacing $y$ with $x$, we obtain: $$f^{-1}(x) = 2 + \sqrt{x-1} \quad \text{for } x \in [1,2)$$ Combining the results from Step 2 and Step 4, the inverse function $f^{-1}(x)$ is defined piecewise as: $$f^{-1}(x) = \begin{cases} x+1 & \text{for } x \in (0,1) \\ 2 + \sqrt{x-1} & \text{for } x \in [1,2) \end{cases}$$