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Let's go through the two parts of the problem.

### Part (a): Finding the domain of \( f(x) = 7 + 5 \arcsin(6x - 4) \)

The function involves an arcsine (\(\arcsin\)) term, which has the domain constraint:
\[
-1 \leq (6x - 4) \leq 1
\]

1. First, solve for \(x\):
   \[
   -1 \leq 6x - 4 \leq 1
   \]

2. Add 4 to all sides:
   \[
   3 \leq 6x \leq 5
   \]

3. Now, divide by 6:
   \[
   \frac{1}{2} \leq x \leq \frac{5}{6}
   \]

Thus, the domain of \( f(x) \) is \( \left[ \frac{1}{2}, \frac{5}{6} \right] \).

### Part (b): Finding the range of \( f(x) = 8 + 2 \arctan(7x + 3) \)

The range of the arctangent (\(\arctan\)) function is \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \). Since this function is modified, we need to adjust the range based on the transformations applied.

1. The arctangent term is multiplied by 2, so the range becomes:
   \[
   2 \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) = \left( -\pi, \pi \right)
   \]

2. Now, adding 8 shifts the range:
   \[
   \left( 8 - \pi, 8 + \pi \right)
   \]

Thus, the range of \( f(x) \) is \( \left( 8 - \pi, 8 + \pi \right) \).

Let me know if you need any clarifications or more details! Here are five related questions to expand your understanding:

1. How does changing the coefficient inside the arcsine or arctangent functions affect the domain or range?
2. What is the difference between the domain constraints of \(\arcsin\) and \(\arctan\)?
3. How would the domain of \(f(x)\) change if it involved an \(\arccos\) function instead of an \(\arcsin\)?
4. What transformations generally affect the range of inverse trigonometric functions?
5. How does shifting a function vertically (like adding 7 or 8) affect the overall range?

**Tip:** Remember that the range of inverse trigonometric functions like arcsine and arctangent is always limited, and understanding their base ranges helps greatly with solving these problems!

Find the domain of f(x) = 7 + 5 arcsin(6x - 4) and the range of f(x) = 8 + 2 arctan(7x + 3).

Learn how to find the domain of f(x) = 7 + 5 arcsin(6x - 4) and the range of f(x) = 8 + 2 arctan(7x + 3). This problem covers key concepts of inverse trigonometric functions, including how transformations affect domain and range. Suitable for students in Grades 11-12.

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