Question

It is given that function \( f(x) = x^4 - 62x^2 + ax + 9 \) attains a local maximum value at \( x = 1 \). Find the value of \( a \), hence obtain all other points where the given function \( f(x) \) attains local maximum or local minimum values.

Hint:

To find critical points, solve \( f'(x) = 0 \). Use the second derivative test to determine the nature of the extrema.

Answer 1

1. Derivative Calculation: The derivative of the function is \( f'(x) = 4x^3 - 124x + a \). 2. Local Maximum Condition: For a local maximum at \( x = 1 \), the first derivative must be zero: \( f'(1) = 4(1)^3 - 124(1) + a = 0 \). Solving for \( a \) yields \( 4 - 124 + a = 0 \), so \( a = -6 \). 3. Function Update: With \( a = -6 \), the function becomes \( f(x) = x^4 - 62x^2 - 6x + 9 \). 4. Identifying Other Critical Points: Critical points occur when \( f'(x) = 0 \). Substituting \( a = -6 \) gives \( 4x^3 - 124x - 6 = 0 \). This simplifies to \( 2x(2x^2 - 62) - 6 = 0 \), further to \( x(2x^2 - 62) = 3 \), and finally to the cubic equation \( 2x^3 - 62x - 3 = 0 \). The roots of this equation can be found numerically; however, \( x = 1 \) is already identified, and symmetry can be considered. 5. Critical Point Analysis: The second derivative is \( f''(x) = 12x^2 - 124 \). At \( x = 1 \), \( f''(1) = 12(1)^2 - 124 = -112 \). Since \( f''(1)<0 \), \( x = 1 \) corresponds to a local maximum. 6. Summary: - A local maximum is confirmed at \( x = 1 \) when \( a = -6 \). - Additional critical points are the roots of the cubic equation \( 2x^3 - 62x - 3 = 0 \), which require numerical approximation. Final Answer: The value of \( a \) is \( \boxed{-6} \). The local maximum occurs at \( x = 1 \).