Question

For what value of x, the function \(f(x) = 2x^3 + 3x^2 - 36x + 10\) has minimum

• A. -2
• B. -3
• C. 2
• D. 1

Hint:

For a cubic polynomial with two critical points, the smaller x-value (more negative) typically corresponds to the local maximum, and the larger x-value corresponds to the local minimum, assuming a positive leading coefficient. This can be a quick check.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the x-coordinate of the point where the given polynomial function has a local minimum value.

Step 2: Key Formula or Approach:
We use the second derivative test to find local minima and maxima.
1. Find the first derivative, \(f'(x)\).
2. Find the critical points by solving the equation \(f'(x) = 0\).
3. Find the second derivative, \(f''(x)\).
4. For each critical point \(c\), evaluate \(f''(c)\).
- If \(f''(c) \gt 0\), the function has a local minimum at \(x=c\).
- If \(f''(c) \lt 0\), the function has a local maximum at \(x=c\).

Step 3: Detailed Explanation:
The function is \(f(x) = 2x^3 + 3x^2 - 36x + 10\).
1. Find the first derivative:
\[ f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 36x + 10) = 6x^2 + 6x - 36 \]
2. Find the critical points by setting \(f'(x) = 0\):
\[ 6x^2 + 6x - 36 = 0 \]
Divide by 6 to simplify:
\[ x^2 + x - 6 = 0 \]
Factor the quadratic equation:
\[ (x+3)(x-2) = 0 \]
The critical points are \(x = -3\) and \(x = 2\).
3. Find the second derivative:
\[ f''(x) = \frac{d}{dx}(6x^2 + 6x - 36) = 12x + 6 \]
4. Apply the second derivative test for each critical point:
For \(x = -3\):
\[ f''(-3) = 12(-3) + 6 = -36 + 6 = -30 \]
Since \(f''(-3) \lt 0\), there is a local maximum at \(x = -3\).
For \(x = 2\):
\[ f''(2) = 12(2) + 6 = 24 + 6 = 30 \]
Since \(f''(2) \gt 0\), there is a local minimum at \(x = 2\).

Step 4: Final Answer:
The function has a minimum at the value \(x = 2\).