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The problem given is:

\[
f(x) = \cos\left(\frac{\pi}{7}\right) + 5x^2 + 3, \quad x \geq 1
\]

You need to find \( f^{-1}(247) \).

To find \( f^{-1}(247) \), we want to find \( x \) such that:

\[
f(x) = 247
\]

So, we set up the equation:

\[
\cos\left(\frac{\pi}{7}\right) + 5x^2 + 3 = 247
\]

Now, let's isolate \( x^2 \):

\[
5x^2 = 247 - \cos\left(\frac{\pi}{7}\right) - 3
\]

\[
5x^2 = 244 - \cos\left(\frac{\pi}{7}\right)
\]

\[
x^2 = \frac{244 - \cos\left(\frac{\pi}{7}\right)}{5}
\]

Finally, take the square root of both sides to find \( x \):

\[
x = \sqrt{\frac{244 - \cos\left(\frac{\pi}{7}\right)}{5}}
\]

Since \( x \geq 1 \) is specified, we only take the positive square root. 

Now let's calculate the numerical value.

The value of \( f^{-1}(247) \) is approximately \( 6.973 \).

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. What is the general process for finding the inverse of a function?
2. How does the domain restriction \( x \geq 1 \) affect the function and its inverse?
3. How would you verify that this \( x \) value is indeed the correct inverse for \( f(247) \)?
4. What would happen if there was no restriction on \( x \) in the function \( f(x) \)?
5. How does the cosine function in \( f(x) \) affect the overall shape of the graph?

**Tip:** Always check the domain and range when working with functions and their inverses to ensure valid solutions.

Let f(x) = cos(π/7) + 5x^2 + 3, x ≥ 1. Find f^(-1)(247).

Explore how to find the inverse of the function f(x) = cos(π/7) + 5x^2 + 3 for x ≥ 1. This step-by-step solution includes key algebraic and trigonometric concepts, such as solving quadratic equations and understanding inverse functions.

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