The problem in the image asks to find the range of the function:

$f(x) = 8 + 2 \cdot \arctan(7x + 3)$

Let's break this down step by step.

Step 1: Understanding the function

The function involves an arctangent term:

$f(x) = 8 + 2 \arctan(7x + 3)$

The arctangent function, $\arctan(x)$, has a well-known range:

$\arctan(x) \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$

Step 2: Scaling the arctangent

Since $\arctan(7x + 3)$ is just a shifted input to the arctangent function, the output range of $\arctan(7x + 3)$ is still the same as for $\arctan(x)$, that is:

$\arctan(7x + 3) \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$

Next, multiply the arctangent by 2:

$2 \cdot \arctan(7x + 3) \in \left( -\pi, \pi \right)$

Step 3: Adding the constant

Now, add 8 to the entire expression:

$f(x) = 8 + 2 \cdot \arctan(7x + 3) \in \left( 8 - \pi, 8 + \pi \right)$

Thus, the range of $f(x)$ is:

$(a, b) = \left( 8 - \pi, 8 + \pi \right)$

Step 4: Approximating the values

Since $\pi \approx 3.1416$:

$a = 8 - \pi \approx 8 - 3.1416 = 4.8584$ $b = 8 + \pi \approx 8 + 3.1416 = 11.1416$

Final Answer:

The range of the function $f(x)$ is approximately:

$(4.8584, 11.1416)$

Would you like a detailed explanation of any specific part? Here are 5 related questions to expand on the topic:

1. What is the general behavior of the $\arctan(x)$ function?
2. How does scaling affect the range of a trigonometric function?
3. Why is the range of $\arctan(x)$ limited to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$?
4. How do you handle functions that involve more complex transformations of $x$?
5. How can you derive the inverse function of a trigonometric expression like this?

Tip: For any inverse trigonometric function like $\arctan(x)$, always start by identifying the standard range before applying transformations like scaling and shifting.