Question

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

Hint:

To identify the nature of critical points, use the second derivative test. If \( f''(x)>0 \), it's a local minimum; if \( f''(x)<0 \), it's a local maximum.

Answer 1

Step 1: Calculate the derivative of \( f(x) \) to find critical points. The derivative is: \[ f'(x) = 4x^3 - 124x + a. \] Given that \( f(x) \) has a local maximum at \( x = 1 \), it follows that: \[ f'(1) = 4(1)^3 - 124(1) + a = 0. \] Solving for \( a \): \[ 4 - 124 + a = 0 \implies a = 120. \] Step 2: Determine the critical points. Substitute \( a = 120 \) into \( f'(x) \): \[ f'(x) = 4x^3 - 124x + 120. \] Factorize \( f'(x) \): \[ f'(x) = 4(x - 1)(x^2 + x - 30). \] Further factorization yields: \[ f'(x) = 4(x - 1)(x - 5)(x + 6). \] The critical points are \( x = -6, 1, 5 \).
Step 3: Classify the critical points using the second derivative \( f''(x) \). The second derivative is: \[ f''(x) = 12x^2 - 124. \] Evaluate \( f''(x) \) at each critical point: \[ f''(-6) = 12(-6)^2 - 124 = 432 - 124 = 308 > 0 \quad (\text{local minimum at } x = -6). \] \[ f''(1) = 12(1)^2 - 124 = 12 - 124 = -112 < 0 \quad (\text{local maximum at } x = 1). \] \[ f''(5) = 12(5)^2 - 124 = 300 - 124 = 176 > 0 \quad (\text{local minimum at } x = 5). \] 
Conclusion:
The function \( f(x) \) has: \[ \text{A local maximum at } x = 1, \quad \text{and local minima at } x = -6, 5. \]