Questions: Let f be the function defined by f(x) = sin x + cos x. What is the absolute minimum value of f on the interval [0,2 π] ? (A) -2 (B) -√2 (C) -1 (D) 0

Transcript text: Let $f$ be the function defined by $f(x)=\sin x+\cos x$. What is the absolute minimum value of $f$ on the interval $[0,2 \pi]$ ? (A) -2 (B) $-\sqrt{2}$ (C) -1 (D) 0

Solution

Solution Steps

To find the absolute minimum value of the function $ f(x) = \sin x + \cos x $ on the interval $[0, 2\pi]$, we can use calculus. First, find the derivative of $ f(x) $ and set it to zero to find critical points. Then, evaluate $ f(x) $ at these critical points and the endpoints of the interval to determine the minimum value.

Step 1: Define the Function and Derivative

The function is given by $ f(x) = \sin x + \cos x $. To find critical points, we first compute the derivative:

\[ f'(x) = \cos x - \sin x \]

Step 2: Find Critical Points

Set the derivative to zero to find critical points:

\[ \cos x - \sin x = 0 \implies \tan x = 1 \]

The solutions to $\tan x = 1$ in the interval $[0, 2\pi]$ are:

\[ x = \frac{\pi}{4}, \frac{5\pi}{4} \]

Step 3: Evaluate the Function at Critical Points and Endpoints

Evaluate $ f(x) $ at the critical points and the endpoints of the interval $[0, 2\pi]$:

$ f(0) = \sin 0 + \cos 0 = 1 $
$ f(2\pi) = \sin 2\pi + \cos 2\pi = 1 $
$ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \sqrt{2} $
$ f\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) = -\sqrt{2} $

Step 4: Determine the Minimum Value

Compare the values obtained: