Step 1: Understanding the Concept:
To find the local minimum or maximum of a function, we use the first and second derivative tests. A local minimum occurs at a critical point where the function's slope changes from negative to positive, which corresponds to a positive second derivative.
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, $f'(x)$.
2. Find the critical points by solving the equation $f'(x) = 0$.
3. Find the second derivative of the function, $f''(x)$.
4. Evaluate $f''(x)$ at each critical point.
- If $f''(c)>0$, the function has a local minimum at $x=c$.
- If $f''(c)<0$, the function has a local maximum at $x=c$.
Step 3: Detailed Explanation:
The given 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. Set the first derivative to zero to find critical points:
\[ 6x^2 + 6x - 36 = 0 \]
Divide by 6:
\[ x^2 + x - 6 = 0 \]
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 to each critical point:
- For $x = -3$:
\[ f''(-3) = 12(-3) + 6 = -36 + 6 = -30 \]
Since $f''(-3)<0$, the function has a local maximum at $x = -3$.
- For $x = 2$:
\[ f''(2) = 12(2) + 6 = 24 + 6 = 30 \]
Since $f''(2)>0$, the function has a local minimum at $x = 2$.
Step 4: Final Answer:
The function has a minimum at $x=2$. Therefore, option (C) is correct.