To find the two values of $c$ in $\left( -\frac{\pi}{4}, \frac{\pi}{4} \right)$ such that $f(c)$ equals the average value of the function $f(x) = 4 \cos(2x)$ on the interval $\left( -\frac{\pi}{4}, \frac{\pi}{4} \right)$, we need to follow these steps:

Step 1: Find the average value of $f(x)$ on the interval $\left( -\frac{\pi}{4}, \frac{\pi}{4} \right)$.

The average value of a function $f(x)$ on the interval $[a, b]$ is given by the formula:

$\text{Average value of } f(x) = \frac{1}{b - a} \int_a^b f(x) \, dx$

For $f(x) = 4 \cos(2x)$ and the interval $\left( -\frac{\pi}{4}, \frac{\pi}{4} \right)$, we have $a = -\frac{\pi}{4}$ and $b = \frac{\pi}{4}$. So the average value becomes:

$\text{Average value of } f(x) = \frac{1}{\frac{\pi}{4} - \left( -\frac{\pi}{4} \right)} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 4 \cos(2x) \, dx$

Simplifying the interval length:

$\text{Average value of } f(x) = \frac{1}{\frac{\pi}{2}} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 4 \cos(2x) \, dx$

Step 2: Compute the integral of $4 \cos(2x)$.

To solve this integral, we use the fact that the integral of $\cos(2x)$ is $\frac{1}{2} \sin(2x)$. So:

$\int 4 \cos(2x) \, dx = 2 \sin(2x)$

Now, evaluating the definite integral:

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 4 \cos(2x) \, dx = 2 \sin\left( 2 \cdot \frac{\pi}{4} \right) - 2 \sin\left( 2 \cdot \left( -\frac{\pi}{4} \right) \right)$

$= 2 \sin\left( \frac{\pi}{2} \right) - 2 \sin\left( -\frac{\pi}{2} \right)$

$= 2(1) - 2(-1) = 2 + 2 = 4$

Step 3: Calculate the average value.

Now substitute this result into the formula for the average value:

$\text{Average value of } f(x) = \frac{1}{\frac{\pi}{2}} \times 4 = \frac{8}{\pi}$

$\text{Average value of } f(x) \approx \frac{8}{3.1416} \approx 2.546$

Step 4: Find the values of $c$ such that $f(c) = \text{Average value of } f(x)$.

Now we need to find $c$ such that:

$f(c) = 4 \cos(2c) = 2.546$

Solve for $c$:

$\cos(2c) = \frac{2.546}{4} = 0.6365$

Now take the inverse cosine:

$2c = \cos^{-1}(0.6365)$

Using a calculator:

$2c \approx 0.876$

So:

$c \approx \frac{0.876}{2} = 0.438$

Thus, one solution is $c \approx 0.438$.

Since the cosine function is periodic, we also have another solution within the given interval. The second solution is:

$2c = -0.876 \quad \Rightarrow \quad c \approx \frac{-0.876}{2} = -0.438$

Thus, the two values of $c$ are approximately:

$c \approx 0.438, -0.438$

Final Answer:

The two values of $c$ are:

$\boxed{0.438, -0.438}$

Let me know if you'd like further clarification or if you have any other questions!