Let `f(x)={(x,-1lexle1),(x^(2),1ltxle2):}` the range of `h^(-1)(x)`, where `h(x)=fof(x)` is (A) `[-1,sqrt(2)]` (B) `[-1,2]` (C) `[-1,4]` (D) `[-2,2]`

Options

A. `[-1,sqrt(2)]`
B. `[-1,2]`
C. `[-1,4]`
D. `[-2,2]`

Explanation

To find the range of \( h^{-1}(x) \) where \( h(x) = f(f(x)) \) and \( f(x) \) is defined as follows:

\[ f(x) = \begin{cases} x & \text{for } -1 \leq x \leq 1 \\ x^2 & \text{for } 1 < x \leq 2 \end{cases} \]

### Step 1: Determine the range of \( f(x) \)

1. For \( -1 \leq x \leq 1 \), \( f(x) = x \). Thus, the range of \( f(x) \) in this interval is \([-1, 1]\).
2. For \( 1 < x \leq 2 \), \( f(x) = x^2 \). The minimum value occurs at \( x = 1 \) (where \( f(1) = 1^2 = 1 \)) and the maximum value occurs at \( x = 2 \) (where \( f(2) = 2^2 = 4 \)). Thus, the range of \( f(x) \) in this interval is \((1, 4]\).

Combining both ranges, the overall range of \( f(x) \) is:

\[ [-1, 1] \cup (1, 4] = [-1, 4] \]

### Step 2: Determine \( h(x) = f(f(x)) \)

Next, we need to evaluate \( h(x) = f(f(x)) \) for different intervals of \( x \).

1. **For \( -1 \leq x \leq 1 \)**:
- Here, \( f(x) = x \).
- Therefore, \( h(x) = f(f(x)) = f(x) = x \).
- The range of \( h(x) \) in this interval is \([-1, 1]\).

2. **For \( 1 < x \leq 2 \)**:
- Here, \( f(x) = x^2 \).
- Since \( x \) in this interval ranges from just above 1 to 2, \( f(x) \) will range from just above 1 to 4.
- We need to evaluate \( f(f(x)) = f(x^2) \):
- \( x^2 \) will be in the interval \((1, 4]\).
- For \( 1 < x^2 \leq 4 \), we need to consider two cases:
- If \( 1 < x^2 \leq 1 \) (which is not possible), we would use \( f(x^2) = x^2 \).
- If \( 1 < x^2 \leq 4 \), we have \( f(x^2) = (x^2)^2 = x^4 \).
- The minimum value of \( x^4 \) occurs at \( x = \sqrt{1} = 1 \) (where \( f(1) = 1^4 = 1 \)) and the maximum value occurs at \( x = 2 \) (where \( f(2) = 2^4 = 16 \)).
- Thus, the range of \( h(x) \) in this interval is \((1, 16]\).

### Step 3: Combine the ranges from both intervals

Combining the ranges from both intervals, we have:

- From \( -1 \leq x \leq 1 \): Range is \([-1, 1]\)
- From \( 1 < x \leq 2 \): Range is \((1, 16]\)

Thus, the overall range of \( h(x) \) is:

\[ [-1, 16] \]

### Step 4: Find the range of \( h^{-1}(x) \)

Since \( h(x) \) maps to the interval \([-1, 16]\), the range of \( h^{-1}(x) \) will be the same as the domain of \( h(x) \):

- The minimum value of \( h^{-1}(x) \) occurs at \( -1 \) (where \( h(-1) = -1 \)).
- The maximum value of \( h^{-1}(x) \) occurs at \( 16 \) (where \( h(2) = 16 \)).

Thus, the range of \( h^{-1}(x) \) is:

\[ [-1, 4] \]

### Conclusion

The correct answer is:

**(C) \([-1, 4]\)**