To find the absolute maximum and minimum of the function
$f(x) = x - \tan x$ on the closed interval $\left[ -\frac{\pi}{4}, \frac{\pi}{4} \right]$, follow these steps:

Step 1: Compute the Derivative

Differentiate $f(x)$:

$f'(x) = 1 - \sec^2 x$

Using the identity $\sec^2 x = 1 + \tan^2 x$, we get:

$f'(x) = 1 - (1 + \tan^2 x) = -\tan^2 x$

Since $\tan^2 x \geq 0$, we see that:

$f'(x) \leq 0$

which means $f(x)$ is decreasing on the entire interval.

Step 2: Find Critical Points

To find critical points, set $f'(x) = 0$:

$-\tan^2 x = 0$

$\tan^2 x = 0$

$\tan x = 0$

This happens at $x = 0$, which is a critical point.

Step 3: Evaluate at Critical Points and Endpoints

Find $f(x)$ at the endpoints and at $x = 0$:

1. $f\left(-\frac{\pi}{4}\right) = -\frac{\pi}{4} - \tan\left(-\frac{\pi}{4}\right)$
   $= -\frac{\pi}{4} - (-1) = -\frac{\pi}{4} + 1$

2. $f(0) = 0 - \tan(0) = 0$

3. $f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} - \tan\left(\frac{\pi}{4}\right)$
   $= \frac{\pi}{4} - 1$

Step 4: Determine Maximum and Minimum

• $f\left(-\frac{\pi}{4}\right) = -\frac{\pi}{4} + 1$
• $f(0) = 0$
• $f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} - 1$

Since $f(x)$ is decreasing, the maximum value is at $x = -\frac{\pi}{4}$, and the minimum value is at $x = \frac{\pi}{4}$.

Final Answer:

• Absolute maximum: $f(-\frac{\pi}{4}) = 1 - \frac{\pi}{4}$
• Absolute minimum: $f(\frac{\pi}{4}) = \frac{\pi}{4} - 1$

Would you like a graphical explanation for better understanding?

Related Questions:

1. How do you determine if a function is increasing or decreasing using derivatives?
2. What is the significance of critical points in finding extrema?
3. How do you find the absolute extrema of a function on a closed interval?
4. How does the tangent function behave within the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$?
5. Why do we evaluate endpoints when finding absolute extrema?

Tip:

When a function is always decreasing (or always increasing) on an interval, the maximum must be at the left endpoint and the minimum at the right endpoint.