Question

For the function \(f(x) = 2x^3 - 15x^2 + 36x + 10\), the local maxima and local minima occur respectively at:

• A. \(x = 3\) and \(x = 2\)
• B. \(x = 1\) and \(x = 3\)
• C. \(x = 2\) and \(x = 3\)
• D. \(x = 3\) and \(x = 4\)

Hint:

To find local maxima and minima, use the first and second derivative tests. Critical points where \(f'(x) = 0\) are candidates for extrema.

Answer 1

The Correct Option is C

Step 1: Calculate the first derivative. To identify potential local maxima and minima, compute the first derivative of the function: \[ f'(x) = 6x^2 - 30x + 36 \]

Step 2: Equate the first derivative to zero. To find the critical points, set the first derivative equal to zero: \[ 6x^2 - 30x + 36 = 0 \] Divide by 6: \[ x^2 - 5x + 6 = 0 \] Factor the quadratic equation: \[ (x - 2)(x - 3) = 0 \]

Step 3: Determine the critical points. The critical points are obtained by solving the equation: \(x = 2\) and \(x = 3\).

Step 4: Apply the Second Derivative Test. The second derivative is used to classify these critical points: \[ f''(x) = 12x - 30 \] Evaluate the second derivative at each critical point: - At \(x = 2\), \(f''(2) = 12(2) - 30 = -6\). A negative second derivative indicates a local maximum. - At \(x = 3\), \(f''(3) = 12(3) - 30 = 6\). A positive second derivative indicates a local minimum.

Final Answer: \[ \boxed{x = 2 \text{ (local maxima)} \text{ and } x = 3 \text{ (local minima)}} \]