Find the average value of \( f(x)=-\cos x \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \). Draw a graph of the function and indicate the average value. The average value is (Type an exact answer, using \( \pi \) as needed.)

To find the average value of the function \( f(x) = -\cos x \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \), we use the formula for the average value of a function, which is given by: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \] For our problem, \( a = -\frac{\pi}{4} \) and \( b = \frac{\pi}{4} \). Therefore, we can calculate the average value as follows: \[ \text{Average value} = \frac{1}{\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} -\cos x \, dx \] Calculating the length of the interval: \[ b - a = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \] Now, we need to evaluate the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} -\cos x \, dx \): \[ \int -\cos x \, dx = -\sin x + C \] Calculating the definite integral: \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} -\cos x \, dx = \left[-\sin x\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = -\sin\left(\frac{\pi}{4}\right) - \left(-\sin\left(-\frac{\pi}{4}\right)\right) \] Since \( \sin\left(\frac{\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \): \[ = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 \] The definite integral is \( -(\left[-\sin x\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}) = -\left( 0 \right) = 0 \). Now we can find the average value: \[ \text{Average value} = \frac{1}{\frac{\pi}{2}} \cdot 0 = 0 \] Thus, the average value of the function over the specified interval is: \[ \boxed{0} \]