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To solve this problem, we need to apply the Intermediate Value Theorem (IVT). The IVT states that if a function \( f \) is continuous on a closed interval \([a, b]\) and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \) in the interval \((a, b)\) such that \( f(c) = N \).

Given:
\[ f(x) = 2x + \tan\left(\frac{x}{5}\right) - 15 \]
and the interval \([10, 15]\).

Let's evaluate the function at the endpoints of the interval:
1. Calculate \( f(10) \):
   \[
   f(10) = 2(10) + \tan\left(\frac{10}{5}\right) - 15 = 20 + \tan(2) - 15 = 5 + \tan(2)
   \]

2. Calculate \( f(15) \):
   \[
   f(15) = 2(15) + \tan\left(\frac{15}{5}\right) - 15 = 30 + \tan(3) - 15 = 15 + \tan(3)
   \]

Now, let's approximate the values using the known values of \( \tan(2) \) and \( \tan(3) \):
   - \( \tan(2) \approx -2.185 \)
   - \( \tan(3) \approx -0.1425 \)

Substitute these values in:
1. \( f(10) \approx 5 + (-2.185) = 2.815 \)
2. \( f(15) \approx 15 + (-0.1425) = 14.8575 \)

Now, using the IVT:
- Since \( f(10) \approx 2.815 \) and \( f(15) \approx 14.8575 \), \( f(x) \) takes on all values between \( 2.815 \) and \( 14.8575 \) in the interval \([10, 15]\).

The question asks for which equation \( f(x) \) has a solution within this interval. The options are:
- \( f(x) = -15 \)
- \( f(x) = 0 \)
- \( f(x) = 5 \)
- \( f(x) = 15 \)

From the range \( [2.815, 14.8575] \), the value \( f(x) = 5 \) lies within this interval, thus guaranteeing a solution in \( [10, 15] \).

Therefore, the correct answer is:
**C. \( f(x) = 5 \)**

Would you like more details on this problem or have any further questions?

Here are some related questions you might find interesting:
1. How do you prove that a function is continuous on a given interval?
2. What are the conditions for applying the Intermediate Value Theorem?
3. How do you find the approximate value of trigonometric functions like \( \tan(2) \) and \( \tan(3) \)?
4. What happens if the function is not continuous on the interval in the IVT?
5. Can the IVT be applied to functions with discontinuities?

**Tip:** When using the Intermediate Value Theorem, always ensure the function is continuous on the closed interval you are examining.

Let f be the function given by f(x) = 2x + tan(x/5) - 15. The Intermediate Value Theorem applied to f on the closed interval [10, 15] guarantees a solution in [10, 15] to which of the following equations?

Apply the Intermediate Value Theorem to solve the function f(x) = 2x + tan(x/5) - 15 on the interval [10, 15]. This detailed solution covers the function's continuity and shows how to find values within the range of the equation.

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