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The problem given is:

**Let \( f(x) = \cos\left(\frac{\pi x}{7}\right) + 5x^2 + 3 \), where \( x \geq 1 \). Find \( f^{-1}(247) \).**

To solve for \( f^{-1}(247) \), we first interpret this as finding \( x \) such that:

\[
f(x) = 247
\]

### Step 1: Set the function equal to 247
\[
\cos\left(\frac{\pi x}{7}\right) + 5x^2 + 3 = 247
\]

Subtract 3 from both sides:
\[
\cos\left(\frac{\pi x}{7}\right) + 5x^2 = 244
\]

### Step 2: Solve for \( x \)
We can attempt to solve this equation by trial and error or numerically, but for simplicity, let's first try solving it with \( \cos\left(\frac{\pi x}{7}\right) \approx 1 \), which happens when \( x = 0 \mod 7 \) (approximately). Then we can solve the quadratic part.

Let me calculate a possible solution for \( x \).It appears that solving the equation directly in terms of both \( x \) and the cosine function simultaneously is complex algebraically. However, the best approach is

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

Learn how to solve for the inverse of the function f(x) = cos(πx/7) + 5x^2 + 3 for x ≥ 1 and find f^{-1}(247) using a combination of trigonometric identities and quadratic equations.

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