Question

Find the product of the maximum and minimum values of the function \( f(x) = \cos x - \frac{1}{2}\cos 2x,\ x \in \left[0, \frac{\pi}{2}\right] \).

Hint:

Find the critical points from \(f^{\prime}(x) = 0\), compare \(f\) at those points and at the end points, then multiply the largest and smallest values.

Answer 1

Step 1: Understand the question:
We need the largest and the smallest value of \(f(x)\) on the closed interval \([0, \frac{\pi}{2}]\), and then their product.
On a closed interval a continuous function reaches its extreme values either at a critical point or at an end point.

Step 2: Find the critical points:
Differentiate the function.
\[ f^{\prime}(x) = -\sin x + \sin 2x = -\sin x + 2\sin x \cos x = \sin x\,(2\cos x - 1) \]
Put \(f^{\prime}(x) = 0\). Then \(\sin x = 0\) or \(\cos x = \frac{1}{2}\).
In \([0, \frac{\pi}{2}]\) this gives \(x = 0\) and \(x = \frac{\pi}{3}\).

Step 3: Evaluate f at the critical points and the end points:
The points to test are \(x = 0,\ \frac{\pi}{3},\ \frac{\pi}{2}\).
\[ f(0) = \cos 0 - \tfrac{1}{2}\cos 0 = 1 - \tfrac{1}{2} = \tfrac{1}{2} \]
\[ f\left(\tfrac{\pi}{3}\right) = \cos\tfrac{\pi}{3} - \tfrac{1}{2}\cos\tfrac{2\pi}{3} = \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4} \]
\[ f\left(\tfrac{\pi}{2}\right) = \cos\tfrac{\pi}{2} - \tfrac{1}{2}\cos \pi = 0 + \tfrac{1}{2} = \tfrac{1}{2} \]

Step 4: Pick the maximum and the minimum:
The largest of the three values is \(\frac{3}{4}\), so the maximum value is \(\frac{3}{4}\).
The smallest is \(\frac{1}{2}\), so the minimum value is \(\frac{1}{2}\).

Step 5: Take the product:
\[ \text{Product} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \]

Final Answer:
The product of the maximum and minimum values is \(\frac{3}{8}\). \[ \boxed{\frac{3}{8}} \]