Step 1: Understanding the Concept:
Local maximum and minimum points of a smooth continuous function occur at critical locations where the slope curve flattens out to zero ($f'(x) = 0$). To find whether a critical turning point is a local peak (maximum) or a local valley (minimum), we use the Second Derivative Test:
- If $f''(c)<0$ at a critical point, the graph curves downward (concave down), meaning $x = c$ is a local maximum.
- If $f''(c)>0$ at a critical point, the graph curves upward (concave up), meaning $x = c$ is a local minimum.
Step 2: Detailed Explanation:
Let's find the stationary points by applying basic calculus optimization rules step-by-step:
1. Step 1: Find the first derivative $f'(x)$ and solve for the critical points:
Given the function: $f(x) = -x^3 + 3x + 1$
Take the derivative with respect to $x$ using the power rule:
$$ f'(x) = -3x^2 + 3 $$
Set the first derivative to zero to locate the stationary critical points:
$$ -3x^2 + 3 = 0 \implies 3x^2 = 3 \implies x^2 = 1 $$
$$ x = 1 \quad \text{and} \quad x = -1 $$
2. Step 2: Find the second derivative $f''(x)$ to determine concavity:
Differentiate $f'(x) = -3x^2 + 3$ one more time:
$$ f''(x) = -6x $$
3. Step 3: Test our critical points inside the second derivative:
- Test $x = -1$:
$$ f''(-1) = -6 \times (-1) = +6 $$
Since $f''(-1)>0$, the function is concave up at this point, making $x = -1$ a local minimum point.
- Test $x = 1$:
$$ f''(1) = -6 \times (1) = -6 $$
Since $f''(1)<0$, the function is concave down at this point, making $x = 1$ a local maximum point.
Therefore, the function achieves its local maximum at $x = 1$. This matches option (A).
Step 3: Final Answer:
The local maximum point of the function is x = 1.