Question

Global maximum value of the function \( f(x) = \sin x + \cos x, \quad x \in [0, \pi] \) is: ...

Hint:

To find the maximum value of a function, consider rewriting it using trigonometric identities and finding critical points within the given domain.

Answer 1

Step 1: Expressing the function in an alternate form. The given function is: \[ f(x) = \sin x + \cos x. \] We rewrite it using the identity: \[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right). \] Thus, the function can be rewritten as: \[ f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right). \] Step 2: Determining the maximum value. Since the maximum value of \( \sin \theta \) is \( 1 \), the maximum value of \( f(x) \) is: \[ \sqrt{2} \times 1 = \sqrt{2}. \] This maximum is attained when: \[ \sin \left( x + \frac{\pi}{4} \right) = 1. \] Step 3: Finding \( x \) in the given domain. Solving for \( x \): \[ x + \frac{\pi}{4} = \frac{\pi}{2}. \] \[ x = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \] Since \( x = \frac{\pi}{4} \) lies within the given domain \( [0, \pi] \), the global maximum value is confirmed as \( \sqrt{2} \). Conclusion: The global maximum value of \( f(x) \) in the given interval is \( \mathbf{\sqrt{2}} \).