Question

In Exercises \(11-18,\) use analytic methods to find the extreme values of the function on the interval and where they occur. $$f(x)=\sin \left(x+\frac{\pi}{4}\right), \quad 0 \leq x \leq \frac{7 \pi}{4}$$

Short answer

The maximum value of the function on the interval is \(\sqrt{2}\) and it occurs at \(x = \frac{\pi}{4}\) and \(x = \frac{9\pi}{4}\). The minimum value is \(-\sqrt{2}\) and it occurs at \(x = \frac{5\pi}{4}\) and \(x = \frac{13\pi}{4}\).

Step-by-step solution

Find the Derivative

Apply the Chain Rule to determine the derivative of the function: \(f'(x)=\cos \left(x + \frac{\pi}{4}\right)\). This helps identify the critical points where the function may have maximum or minimum values.

Set the derivative to zero

Solve the equation \(f'(x) = 0\) for the interval given. The solutions to this equation are the critical points where the function may achieve its extreme values.

Determine critical points

The critical points occur when \(x = \frac{\pi}{2} - \frac{\pi}{4}\) and \(x = \frac{5\pi}{2} - \frac{\pi}{4}\) within the interval \(0 \leq x \leq \frac{7 \pi}{4}\). These critical points are obtained when the cosine function equals zero.

Evaluate the function at the critical points and endpoints

Evaluate \(f(x)\) at the critical points \(x = \frac{\pi}{4}\) and \(x = \frac{9\pi}{4}\), and at the endpoints \(x = 0\) and \(x = \frac{7\pi}{4}\). Determine which of them yields the maximum and minimum values of the function.

Identify the extreme values

By comparing all the values from Step 4, the highest value is the maximum and the lowest value is the minimum of the function on the interval \(0 \leq x \leq \frac{7\pi}{4}\).