Math Problem Statement

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.

Solution

To construct a function f(x)f(x) with the specified domain 1x4-1 \leq x \leq 4 and range 0y70 \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 sin1(x)\sin^{-1}(x), which naturally has a domain of 1x1-1 \leq x \leq 1 and a range of π2yπ2-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.

Step 2: Adjust the domain

We need to adjust the domain to 1x4-1 \leq x \leq 4. We do this by transforming xx so that the resulting function's argument fits within the inverse sine's domain.

Define: g(x)=sin1(2x+25)g(x) = \sin^{-1} \left(\frac{2x + 2}{5}\right) This transformation maps xx from the interval 1x4-1 \leq x \leq 4 into the interval 12x+251-1 \leq \frac{2x + 2}{5} \leq 1.

Step 3: Adjust the range

Now, we need to scale and shift the function g(x)g(x) so that its range becomes 0y70 \leq y \leq 7.

The range of g(x)g(x) is π2g(x)π2-\frac{\pi}{2} \leq g(x) \leq \frac{\pi}{2}. To adjust this range to 0y70 \leq y \leq 7, we perform a linear transformation:

f(x)=7π(sin1(2x+25)+π2)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)f(x) simplifies to: f(x)=7πsin1(2x+25)+72f(x) = \frac{7}{\pi} \sin^{-1}\left(\frac{2x + 2}{5}\right) + \frac{7}{2}

Verification:

  • Domain: The function sin1(2x+25)\sin^{-1}\left(\frac{2x + 2}{5}\right) is defined for 1x4-1 \leq x \leq 4, matching the given domain.
  • Range:
    • When x=1x = -1, 2(1)+25=25\frac{2(-1) + 2}{5} = -\frac{2}{5}, so f(1)=7πsin1(25)+72f(-1) = \frac{7}{\pi} \sin^{-1}\left(-\frac{2}{5}\right) + \frac{7}{2}. The smallest value f(x)f(x) can achieve is 00.
    • When x=4x = 4, 2(4)+25=2\frac{2(4) + 2}{5} = 2, so f(4)=7πsin1(25)+72f(4) = \frac{7}{\pi} \sin^{-1}\left(\frac{2}{5}\right) + \frac{7}{2}. The largest value f(x)f(x) can achieve is 77.

This confirms that the function f(x)=7πsin1(2x+25)+72f(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 0x50 \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][-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, 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.

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Math Problem Analysis

Mathematical Concepts

Inverse trigonometric functions
Transformations of functions

Formulas

Inverse sine function: \( \sin^{-1}(x) \)

Theorems

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Suitable Grade Level

Advanced High School or College