Finding the Absolute Maximum and Minimum Values of $f(x) = \sin x + \cos x$ on $[0,\pi]$

This solution explains how to find the absolute maximum and minimum values of the function $f(x) = \sin x + \cos x$ within the specific interval $[0, \pi]$. We will use calculus, specifically by finding the function's derivative and evaluating the function at critical points and interval endpoints.

Understanding Absolute Extrema

The absolute maximum value is the largest value the function takes on the interval, and the absolute minimum value is the smallest value. According to the Extreme Value Theorem, a continuous function on a closed interval will always attain both an absolute maximum and an absolute minimum value within that interval. These extreme values occur either at the critical points of the function (where the derivative is zero or undefined) or at the endpoints of the interval.

Step 1: Differentiate the Function

First, we find the derivative of the function $f(x) = \sin x + \cos x$ with respect to $x$. This helps us locate the critical points.

The derivative, $f'(x)$, is calculated as:

Step 2: Find Critical Points

Critical points occur where the derivative $f'(x)$ is equal to zero or is undefined. Since $\cos x - \sin x$ is defined for all real numbers, we only need to find where it equals zero.

Set $f'(x) = 0$: $\cos x - \sin x = 0$ $\cos x = \sin x$ To solve for $x$, we can divide both sides by $\cos x$ (assuming $\cos x \neq 0$. If $\cos x = 0$, then $x = \pi/2$ in the interval $[0, \pi]$, and $\sin(\pi/2) = 1 \neq 0$, so $\cos x$ cannot be 0 here). $\frac{\sin x}{\cos x} = 1$ $\tan x = 1$ Now, we need to find the value(s) of $x$ in the interval $[0, \pi]$ for which $\tan x = 1$. The tangent function is positive in the first quadrant. The principal value is $x = \frac{\pi}{4}$. This value, $x = \frac{\pi}{4}$, lies within our interval $[0, \pi]$. So, $x = \frac{\pi}{4}$ is a critical point.

Step 3: Evaluate the Function at Critical Points and Endpoints

We need to evaluate the original function $f(x) = \sin x + \cos x$ at the critical point found and at the endpoints of the interval $[0, \pi]$.

• At the critical point $x = \frac{\pi}{4}$: $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)$ $f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$ $f\left(\frac{\pi}{4}\right) = \sqrt{2}$
• At the left endpoint $x = 0$: $f(0) = \sin(0) + \cos(0)$ $f(0) = 0 + 1$ $f(0) = 1$
• At the right endpoint $x = \pi$: $f(\pi) = \sin(\pi) + \cos(\pi)$ $f(\pi) = 0 + (-1)$ $f(\pi) = -1$

Step 4: Determine the Absolute Maximum and Minimum Values

Compare the function values obtained in the previous step:

• $f(\frac{\pi}{4}) = \sqrt{2}$
• $f(0) = 1$
• $f(\pi) = -1$

The largest value among these is $\sqrt{2}$, and the smallest value is -1.

• The absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$ is $\sqrt{2}$.
• The absolute minimum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$ is -1.

Therefore, the absolute maximum value is $\sqrt{2}$ and the absolute minimum value is -1.