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Part 1
The graph of the function ​f(x)equalsseven eighths sine x
​, minusStartFraction pi Over 2 EndFraction
less than or equalsxless than or equalsStartFraction pi Over 2 EndFraction
 is shown. Use symmetry with respect to the line yequalsx to add the graph of f Superscript negative 1 to the same graph.​ (It is not necessary to find a formula for f Superscript negative 1​.) Identify the domain and range of f Superscript negative 1.

To approach this problem, let's break it down step by step. We're given the function:

\[
f(x) = \frac{7}{8} \sin(x), \quad -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}
\]

We're asked to:
1. Use symmetry with respect to the line \( y = x \) to sketch the inverse function \( f^{-1}(x) \).
2. Identify the domain and range of \( f^{-1}(x) \).

### Step 1: Symmetry and the Graph of \( f^{-1}(x) \)

The inverse of a function \( f(x) \) is the reflection of its graph over the line \( y = x \). This means that for every point \( (x, y) \) on the graph of \( f(x) \), there is a corresponding point \( (y, x) \) on the graph of \( f^{-1}(x) \). 

#### Graph of \( f(x) \)
- The function \( f(x) = \frac{7}{8} \sin(x) \) is a scaled sine function, where the amplitude is \( \frac{7}{8} \) and the domain is restricted to \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). 
- The sine function is symmetric around the origin, meaning \( \sin(-x) = -\sin(x) \), and the graph of \( f(x) \) will reflect this symmetry in the given interval.

#### Graph of \( f^{-1}(x) \)
To obtain the graph of the inverse, simply reflect the graph of \( f(x) \) over the line \( y = x \). If the point \( (x_1, y_1) \) is on the graph of \( f(x) \), then the point \( (y_1, x_1) \) will be on the graph of \( f^{-1}(x) \).

### Step 2: Domain and Range of \( f^{-1}(x) \)

The domain of \( f^{-1}(x) \) is the range of \( f(x) \), and the range of \( f^{-1}(x) \) is the domain of \( f(x) \). 

- **Domain of \( f(x) \):** The domain of \( f(x) \) is \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \).
- **Range of \( f(x) \):** Since \( f(x) = \frac{7}{8} \sin(x) \) and \( \sin(x) \) takes values between -1 and 1, the range of \( f(x) \) will be \( -\frac{7}{8} \leq f(x) \leq \frac{7}{8} \).

Therefore:
- The **domain** of \( f^{-1}(x) \) is \( -\frac{7}{8} \leq x \leq \frac{7}{8} \).
- The **range** of \( f^{-1}(x) \) is \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).

### Conclusion
1. The graph of \( f^{-1}(x) \) is a reflection of the graph of \( f(x) \) over the line \( y = x \).
2. The domain of \( f^{-1}(x) \) is \( -\frac{7}{8} \leq x \leq \frac{7}{8} \), and its range is \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).

Would you like to discuss the graphical approach further or get into more details?

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Here are five related questions to expand on this topic:
1. How would the domain and range of \( f^{-1}(x) \) change if the amplitude of \( f(x) \) were increased?
2. What properties must a function have for it to have an inverse?
3. How can you verify that two functions are inverses of each other?
4. How does the reflection of a function over the line \( y = x \) relate to its derivative?
5. Can a function with a restricted domain still have an inverse over that interval?

**Tip:** Always check the original function for one-to-one behavior (monotonicity) before concluding that an inverse exists.

The graph of the function f(x) = seven eighths sine x, -pi/2 ≤ x ≤ pi/2 is shown. Use symmetry with respect to the line y = x to add the graph of f⁻¹ to the same graph. Identify the domain and range of f⁻¹.

Learn how to graph the inverse of the function f(x) = (7/8)sin(x) over the interval -pi/2 ≤ x ≤ pi/2 by using symmetry across the line y = x. Identify the domain and range of the inverse function f⁻¹(x) and explore key trigonometric concepts and theorems. This problem is ideal for high school students in Grades 10-12.

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