https://file.aiquickdraw.com/mathbot/file/1727319013628vvkjef99.jpg

The problem in the image asks to apply the **Intermediate Value Theorem (IVT)** to the function:

\[
f(x) = \frac{9 + 2xe^{-\frac{x}{4}}}{\cos\left(\frac{x}{2}\right)}
\]

on the closed interval \([24, 28]\), and determine which of the following equations the theorem guarantees a solution for:

- \( A: f(x) = 0 \)
- \( B: f(x) = 9.090 \)
- \( C: f(x) = 12.235 \)
- \( D: f(x) = 76.999 \)

### Steps to approach the problem:

1. **Use of the Intermediate Value Theorem (IVT)**:
   The IVT guarantees that if \(f(x)\) is continuous on a closed interval \([a, b]\) and takes values \(f(a)\) and \(f(b)\) at the endpoints, then for any value \(y\) between \(f(a)\) and \(f(b)\), there is at least one value \(c \in [a, b]\) such that \(f(c) = y\).

2. **Check continuity of \(f(x)\)**:
   The function \(f(x)\) appears to be continuous on \([24, 28]\) as it is composed of elementary continuous functions (assuming the cosine term does not produce any discontinuities in this range, i.e., its denominator does not vanish).

3. **Evaluate \(f(24)\) and \(f(28)\)**:
   We need to compute the values of \(f(x)\) at the endpoints \(x = 24\) and \(x = 28\). Once we have these values, we check which values between \(f(24)\) and \(f(28)\) correspond to the given choices (A, B, C, or D).

Let me compute the values of \(f(24)\) and \(f(28)\).The values of the function \( f(x) \) at the endpoints are:

- \( f(24) \approx 10.81 \)
- \( f(28) \approx 66.19 \)

### Applying the Intermediate Value Theorem:
Since \( f(x) \) is continuous on \([24, 28]\) and \( f(24) = 10.81 \) and \( f(28) = 66.19 \), the IVT guarantees that \( f(x) \) must take on all values between 10.81 and 66.19 at least once in this interval.

Now, checking the given options:
- \( A: f(x) = 0 \) — This value is **not** within the range [10.81, 66.19].
- \( B: f(x) = 9.090 \) — This value is **not** within the range [10.81, 66.19].
- \( C: f(x) = 12.235 \) — This value **is** within the range [10.81, 66.19].
- \( D: f(x) = 76.999 \) — This value is **not** within the range [10.81, 66.19].

Thus, the Intermediate Value Theorem guarantees a solution for the equation:

\[
\boxed{C: f(x) = 12.235}
\]

Would you like more details or have any questions?

Here are 5 related questions to expand on this problem:
1. What are the conditions required for the Intermediate Value Theorem to hold?
2. How would discontinuities in \( f(x) \) affect the application of the IVT?
3. Can the IVT be used to guarantee solutions outside of the function's range on an interval?
4. What other theorems can be applied in similar scenarios to determine the existence of solutions?
5. How can the behavior of \( f(x) \) be further analyzed to understand its trend on the interval?

**Tip:** The Intermediate Value Theorem doesn't provide the exact solution but only guarantees the existence of a solution within the specified range.

Let f(x) = (9 + 2xe^(-x/4)) / cos(x/2) on the interval [24, 28]. Which of the following equations does the Intermediate Value Theorem guarantee a solution for? A: f(x) = 0, B: f(x) = 9.090, C: f(x) = 12.235, D: f(x) = 76.999.

Learn how the Intermediate Value Theorem guarantees a solution to f(x) = (9 + 2xe^(-x/4)) / cos(x/2) on the interval [24, 28]. Detailed explanation provided with step-by-step solution, calculating the function's values at x = 24 and x = 28. Suitable for calculus students.

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