Question

The absolute maximum value of the function f(x)=sinx + cosx, x \(\in\)[0, \(\pi\)] is:

• A. \(\sqrt{2}\)
• B. 2
• C. 1
• D. \(\frac{1}{\sqrt{2}}\)

Answer 1

The Correct Option is A

To find the absolute maximum value of the function \(f(x)=\sin x + \cos x\) on the interval \([0, \pi]\), we will follow these steps:

1. Determine the derivative of the function: \[f'(x)=\cos x - \sin x\]
2. Find the critical points by solving \(f'(x)=0\): \[\cos x - \sin x = 0 \Rightarrow \cos x = \sin x\] Solving \(\cos x = \sin x\), we get: \[\tan x = 1 \Rightarrow x = \frac{\pi}{4}\]
3. Evaluate \(f(x)\) at critical points and endpoints of the interval \([0, \pi]\):
   • At \(x=0\): \[f(0)=\sin 0 + \cos 0 = 0 + 1 = 1\]
   • At \(x=\frac{\pi}{4}\): \[f\left(\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)+\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}\]
   • At \(x=\pi\): \[f(\pi)=\sin \pi + \cos \pi = 0 - 1 = -1\]
4. Compare these values to find the absolute maximum: The values of \(f(x)\) are 1, \(\sqrt{2}\), and -1. Therefore, the absolute maximum value in the interval \([0, \pi]\) is \(\sqrt{2}\).

Thus, the absolute maximum value of \(f(x)=\sin x + \cos x\) on \([0, \pi]\) is \(\sqrt{2}\).