1. Derivative Calculation: The derivative of the function is computed as \(f'(x) = 4x^3 - 124x + a\).
2. Local Maximum Condition: For a local maximum at \(x = 1\), the first derivative must be zero at this point. Substituting \(x = 1\) into \(f'(x)\) yields \(4(1)^3 - 124(1) + a = 0\). This simplifies to \(4 - 124 + a = 0\), which gives \(a = -6\).
3. Function Update: With \(a = -6\), the function becomes \(f(x) = x^4 - 62x^2 - 6x + 9\).
4. Identification of Other Critical Points: To find other critical points, set \(f'(x) = 0\): \(4x^3 - 124x - 6 = 0\). This can be expressed as \(2x(2x^2 - 62) - 6 = 0\). Further simplification leads to \(x(2x^2 - 62) = 3\), and ultimately \(2x^3 - 62x - 3 = 0\). The roots of this cubic equation require numerical methods for precise determination, beyond considering \(x = 1\) and symmetry.
5. Classification of Critical Points: The second derivative is calculated as \(f''(x) = 12x^2 - 124\). Evaluating at \(x = 1\): \(f''(1) = 12(1)^2 - 124 = -112\). Since \(f''(1)<0\), \(x = 1\) corresponds to a local maximum.
6. Summary of Findings: A local maximum is confirmed at \(x = 1\) when \(a = -6\). Additional critical points are roots of the cubic equation \(2x^3 - 62x - 3 = 0\), which are approximated.
Final Answer: The value of \(a\) is \( \boxed{-6} \). The local maximum occurs at \( x = 1 \).