[0,9π/2] to Reals, h(x)=1/2xcos(X), find the maximum 

To find the maximum of the function \( h(x) = \frac{1}{2} x \cos(x) \) over the interval \([0, \frac{9\pi}{2}]\), we can follow these steps:

1. **Define the function**: \( h(x) = \frac{1}{2} x \cos(x) \).

2. **Find the derivative**: To locate the critical points, we'll differentiate \( h(x) \) with respect to \( x \).

\[
h'(x) = \frac{d}{dx} \left( \frac{1}{2} x \cos(x) \right)
\]

Using the product rule, we have:

\[
h'(x) = \frac{1}{2} \left( \cos(x) - x \sin(x) \right)
\]

3. **Set the derivative equal to zero**: Solve \( h'(x) = 0 \) to find the critical points.

\[
\cos(x) - x \sin(x) = 0
\]

\[
\cos(x) = x \sin(x)
\]

\[
\tan(x) = \frac{1}{x}
\]

4. **Check endpoints and critical points**: Evaluate \( h(x) \) at the critical points found in step 3, and also at the endpoints \( x = 0 \) and \( x = \frac{9\pi}{2} \).

5. **Determine the maximum value**: Compare the values obtained from the critical points and endpoints.

Let's compute the exact values of \( h(x) \) at these points to find the maximum.It seems the solution for critical points is in a form that isn't directly iterable, likely due to the transcendental equation involving \(\tan(x) = \frac{1}{x}\). This kind of equation often does not have straightforward algebraic solutions, so let's approximate the critical points numerically within the given interval \([0, \frac{9\pi}{2}]\).

We'll find approximate solutions for the critical points and then evaluate \(h(x)\) at those points and the endpoints to determine the maximum value. Let's proceed with a numerical approach to handle this.The maximum value of the function \( h(x) = \frac{1}{2} x \cos(x) \) over the interval \([0, \frac{9\pi}{2}]\) is approximately **6.303**, which occurs at \( x \approx 12.645 \).

Here are the evaluated points for clarity:

1. Critical Point at \( x \approx 0.860 \), \( h(x) \approx 0.281 \)
2. Critical Point at \( x \approx 3.426 \), \( h(x) \approx -1.644 \)
3. Critical Point at \( x \approx 6.437 \), \( h(x) \approx 3.181 \)
4. Critical Point at \( x \approx 9.529 \), \( h(x) \approx -4.739 \)
5. Critical Point at \( x \approx 12.645 \), \( h(x) \approx 6.303 \)
6. Endpoints at \( x = 0 \), \( h(x) = 0 \)
7. Endpoint at \( x = \frac{9\pi}{2} \approx 14.137 \), \( h(x) \approx 0 \)

The maximum value occurs at the critical point \( x \approx 12.645 \).

Would you like more details or have any questions?

Here are some related questions:
1. How can we interpret the behavior of \( h(x) \) as \( x \) increases?
2. What is the significance of the cosine term in determining the function's maxima and minima?
3. How can we approximate solutions to equations involving trigonometric and polynomial terms?
4. How does the interval affect the location of maximum points in periodic functions?
5. What tools or methods are best for analyzing non-algebraic critical points numerically?

**Tip:** When finding extrema on an interval, always evaluate the function at both critical points and endpoints to ensure the global maximum or minimum is identified.

[0,9π/2] to Reals, h(x)=1/2xcos(X), find the maximum

Find the maximum of the function h(x) = (1/2)x cos(x) on the interval [0, 9π/2]. This detailed solution covers key calculus concepts, including critical points, the product rule, and transcendental equations. Numerical methods are used to approximate solutions, suitable for university-level calculus students.

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