find $$ \left(f^{-1}\right) $$(2), f(x)= $$ x^3+3\text{ sin x}+2\text{ cos }x\text{ } $$

To find $ (f^{-1})(2) $ for $ f(x) = x^3 + 3 \sin x + 2 \cos x $, we need to determine the value of $ x $ such that $ f(x) = 2 $. This requires solving the equation:

\[ x^3 + 3 \sin x + 2 \cos x = 2 \]

Since this is a transcendental equation, it may not have an analytical solution, and we might need numerical methods to find the value of $ x $.

To proceed, we can use a numerical method such as the Newton-Raphson method to approximate the root. First, we define the function and its derivative:

\[ g(x) = x^3 + 3 \sin x + 2 \cos x - 2 \]
\[ g'(x) = 3x^2 + 3 \cos x - 2 \sin x \]

Next, we use the Newton-Raphson iteration formula:

\[ x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} \]

We start with an initial guess $ x_0 $ and iterate until convergence. Let’s start with an initial guess, for instance, $ x_0 = 1 $:

1. $ x_0 = 1 $
2. $ g(x_0) = 1^3 + 3 \sin(1) + 2 \cos(1) - 2 \approx 1 + 2.5244 + 1.0806 - 2 \approx 2.605 $
3. $ g'(x_0) = 3(1)^2 + 3 \cos(1) - 2 \sin(1) \approx 3 + 1.6204 - 1.6829 \approx 2.9375 $
4. $ x_1 = x_0 - \frac{g(x_0)}{g'(x_0)} \approx 1 - \frac{2.605}{2.9375} \approx 0.1144 $

Next iteration:

1. $ x_1 \approx 0.1144 $
2. $ g(x_1) \approx (0.1144)^3 + 3 \sin(0.1144) + 2 \cos(0.1144) - 2 \approx 0.0015 + 0.343 - 1.9957 - 2 \approx -1.6512 $
3. $ g'(x_1) \approx 3(0.1144)^2 + 3 \cos(0.1144) - 2 \sin(0.1144) \approx 0.0392 + 2.9864 - 0.228 \approx 2.7976 $
4. $ x_2 = x_1 - \frac{g(x_1)}{g'(x_1)} \approx 0.1144 - \frac{-1.6512}{2.7976} \approx 0.7042 $

This iterative process would continue until the value of $ x $ converges to a stable value. 

We can now use a numerical solver in Python to find the precise value. Here is the code to find the solution:

```python
import scipy.optimize as opt
import numpy as np

# Define the function
def f(x):
    return x**3 + 3*np.sin(x) + 2*np.cos(x) - 2

# Use a numerical solver to find the root
root = opt.newton(f, 1)
root
```

Running the above code would give us the value of $ x $ such that $ f(x) = 2 $. This $ x $ is $ (f^{-1})(2) $.

If we assume the iterations converge properly, $ (f^{-1})(2) \approx 0.7042 $. This process should be repeated with higher precision to ensure the exact value.

Would you like more details or have any questions about this process? Here are some more questions you might consider:

1. How to derive the Newton-Raphson method?
2. What are transcendental equations?
3. How to use other numerical methods for solving equations?
4. What is the significance of the derivative in the Newton-Raphson method?
5. How to determine a good initial guess for numerical methods?

**Tip:** Always verify the results of numerical methods by substituting them back into the original equation.

Learn how to find the inverse function value (f^-1)(2) for the function f(x) = x^3 + 3sin(x) + 2cos(x). This involves solving a transcendental equation using the Newton-Raphson method. Detailed step-by-step solution and numerical methods applied. Suitable for advanced college-level students and beyond.

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