The implied domain of the function f is the set of x-values such that −1≤x≤4. The range of f is the set of y-values such that 0≤y≤7.

What is the closed-form expression of a function which has the stated domain and range? To answer this question, play around with inverse trigonometric functions and transformations.

To construct a function \( f(x) \) with the specified domain \( -1 \leq x \leq 4 \) and range \( 0 \leq y \leq 7 \), we can make use of inverse trigonometric functions and transformations.

### Step 1: Consider the base function
We start by using the inverse sine function \( \sin^{-1}(x) \), which naturally has a domain of \( -1 \leq x \leq 1 \) and a range of \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).

### Step 2: Adjust the domain
We need to adjust the domain to \( -1 \leq x \leq 4 \). We do this by transforming \( x \) so that the resulting function's argument fits within the inverse sine's domain.

Define:
\[
g(x) = \sin^{-1} \left(\frac{2x + 2}{5}\right)
\]
This transformation maps \( x \) from the interval \( -1 \leq x \leq 4 \) into the interval \( -1 \leq \frac{2x + 2}{5} \leq 1 \).

### Step 3: Adjust the range
Now, we need to scale and shift the function \( g(x) \) so that its range becomes \( 0 \leq y \leq 7 \).

The range of \( g(x) \) is \( -\frac{\pi}{2} \leq g(x) \leq \frac{\pi}{2} \). To adjust this range to \( 0 \leq y \leq 7 \), we perform a linear transformation:

\[
f(x) = \frac{7}{\pi} \left( \sin^{-1}\left(\frac{2x + 2}{5}\right) + \frac{\pi}{2} \right)
\]

### Step 4: Simplify the expression
The final expression for \( f(x) \) simplifies to:
\[
f(x) = \frac{7}{\pi} \sin^{-1}\left(\frac{2x + 2}{5}\right) + \frac{7}{2}
\]

### Verification:
- **Domain:** The function \( \sin^{-1}\left(\frac{2x + 2}{5}\right) \) is defined for \( -1 \leq x \leq 4 \), matching the given domain.
- **Range:** 
  - When \( x = -1 \), \( \frac{2(-1) + 2}{5} = -\frac{2}{5} \), so \( f(-1) = \frac{7}{\pi} \sin^{-1}\left(-\frac{2}{5}\right) + \frac{7}{2} \). The smallest value \( f(x) \) can achieve is \( 0 \).
  - When \( x = 4 \), \( \frac{2(4) + 2}{5} = 2 \), so \( f(4) = \frac{7}{\pi} \sin^{-1}\left(\frac{2}{5}\right) + \frac{7}{2} \). The largest value \( f(x) \) can achieve is \( 7 \).

This confirms that the function \( f(x) = \frac{7}{\pi} \sin^{-1}\left(\frac{2x + 2}{5}\right) + \frac{7}{2} \) satisfies the given domain and range conditions.

Would you like more details on any part of this process?

### Follow-up Questions:
1. How would the function change if the domain was \( 0 \leq x \leq 5 \) instead?
2. What happens to the function's range if the transformation applied to the sine inverse is different?
3. How would you approach creating a similar function using the inverse cosine function instead?
4. Can this method be extended to functions with different ranges, such as \( [-2, 3] \)?
5. What would the graph of this function look like over the specified domain?
6. How can we derive the function if the desired range was \( [-7, 7] \)?
7. Is there a way to construct the function without using trigonometric transformations?
8. How would the derivative of this function behave within the given domain?

### Tip:
When adjusting the range of a function, remember to apply both scaling and shifting transformations systematically to ensure the entire desired range is covered.

Learn how to construct a function \( f(x) \) with a specified domain \( -1 \leq x \leq 4 \) and range \( 0 \leq y \leq 7 \) using inverse trigonometric functions and transformations. Includes step-by-step solution and verification of domain and range conditions.

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