Key Concepts and Formulas

• Increasing Function: A function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ for all $x$ in that interval.
• Derivative: The derivative $f'(x)$ represents the rate of change of the function $f(x)$.
• Quadratic Inequality: To solve a quadratic inequality of the form $ax^2 + bx + c \ge 0$ or $ax^2 + bx + c \le 0$, find the roots of the quadratic equation $ax^2 + bx + c = 0$ and analyze the sign of the quadratic expression in the intervals determined by the roots.

Step-by-Step Solution

1. Analyze Option (A)

• Function: $f(x) = x^3 - 3x^2 + 3x + 3$
• Step 1: Find the derivative $f'(x)$. We differentiate $f(x)$ with respect to $x$ to find its derivative: $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x + 3) = 3x^2 - 6x + 3$
• Step 2: Factorize $f'(x)$. We factor out a 3: $f'(x) = 3(x^2 - 2x + 1)$ Recognize the perfect square trinomial $(x-1)^2$: $f'(x) = 3(x-1)^2$
• Step 3: Determine when $f'(x) \ge 0$. Since $(x-1)^2$ is always greater than or equal to zero for any real number $x$, and $3$ is a positive constant, $f'(x) = 3(x-1)^2 \ge 0$ for all $x \in (-\infty, \infty)$.
• Comparison: The function is increasing on $(-\infty, \infty)$. The given interval is $(-\infty, \infty)$.
• Conclusion: Option (A) is correctly matched.

2. Analyze Option (B)

• Function: $f(x) = 2x^3 - 3x^2 - 12x + 6$
• Step 1: Find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 6) = 6x^2 - 6x - 12$
• Step 2: Factorize $f'(x)$. Factor out 6: $f'(x) = 6(x^2 - x - 2)$ Factor the quadratic expression: $f'(x) = 6(x-2)(x+1)$
• Step 3: Determine when $f'(x) \ge 0$. We need $6(x-2)(x+1) \ge 0$. Since $6 > 0$, we need $(x-2)(x+1) \ge 0$. This inequality holds when both factors have the same sign (both positive or both negative).
  • Case 1: $x-2 \ge 0$ and $x+1 \ge 0 \Rightarrow x \ge 2$ and $x \ge -1 \Rightarrow x \ge 2$.
  • Case 2: $x-2 \le 0$ and $x+1 \le 0 \Rightarrow x \le 2$ and $x \le -1 \Rightarrow x \le -1$. Thus, $f'(x) \ge 0$ when $x \in (-\infty, -1] \cup [2, \infty)$.
• Comparison: The function is increasing on $(-\infty, -1] \cup [2, \infty)$. The given interval is $[2, \infty)$. Since $[2, \infty)$ is a subset of the increasing interval, the function is indeed increasing on $[2, \infty)$.
• Conclusion: Option (B) is correctly matched.

3. Analyze Option (C)

• Function: $f(x) = 3x^2 - 2x + 1$
• Step 1: Find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(3x^2 - 2x + 1) = 6x - 2$
• Step 2: Determine when $f'(x) \ge 0$. We need $6x - 2 \ge 0$. Add 2 to both sides: $6x \ge 2$ Divide by 6: $x \ge \frac{2}{6}$ $x \ge \frac{1}{3}$ Thus, the function $f(x)$ is increasing on the interval $\left[ \frac{1}{3}, \infty \right)$.
• Comparison: The calculated increasing interval is $\left[ \frac{1}{3}, \infty \right)$. The given interval in the option is $\left( -\infty, \frac{1}{3} \right]$. These two intervals are disjoint except for the point $x=1/3$. In fact, on the interval $\left( -\infty, \frac{1}{3} \right)$, $f'(x) = 6x - 2 < 0$, which means the function is decreasing on this interval.
• Conclusion: The function is increasing on $\left[ \frac{1}{3}, \infty \right)$, but the option states it is increasing on $\left( -\infty, \frac{1}{3} \right]$. This is a mismatch. Therefore, Option (C) is incorrectly matched.

4. Analyze Option (D)

• Function: $f(x) = x^3 + 6x^2 + 6$
• Step 1: Find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x^3 + 6x^2 + 6) = 3x^2 + 12x$
• Step 2: Factorize $f'(x)$. Factor out $3x$: $f'(x) = 3x(x+4)$
• Step 3: Determine when $f'(x) \ge 0$. We need $3x(x+4) \ge 0$. Since $3 > 0$, we need $x(x+4) \ge 0$. This inequality holds when both factors have the same sign.
  • Case 1: $x \ge 0$ and $x+4 \ge 0 \Rightarrow x \ge 0$ and $x \ge -4 \Rightarrow x \ge 0$.
  • Case 2: $x \le 0$ and $x+4 \le 0 \Rightarrow x \le 0$ and $x \le -4 \Rightarrow x \le -4$. Thus, $f'(x) \ge 0$ when $x \in (-\infty, -4] \cup [0, \infty)$.
• Comparison: The function is increasing on $(-\infty, -4] \cup [0, \infty)$. The given interval is $(-\infty, -4)$. Since $(-\infty, -4)$ is a subset of the increasing interval, the function is indeed increasing on $(-\infty, -4)$.
• Conclusion: Option (D) is correctly matched.

Common Mistakes & Tips

• Sign Errors: Be extremely careful when solving inequalities, especially when multiplying or dividing by negative numbers.
• Interval Notation: Pay attention to the difference between open and closed intervals, and use the correct notation.
• Checking Endpoints: When dealing with intervals, always check if the endpoints should be included or excluded based on the inequality $f'(x) \ge 0$ or $f'(x) > 0$.

Summary

By analyzing the derivative of each function and comparing the resulting intervals where the function is increasing with the given intervals, we found that option (C) is incorrectly matched. The function $f(x) = 3x^2 - 2x + 1$ is increasing on $\left[ \frac{1}{3}, \infty \right)$, but the option claims it is increasing on $\left( -\infty, \frac{1}{3} \right]$.

The final answer is \boxed{C}, which corresponds to option (C).