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(A) To determine if a function $f(x)$ is increasing on an interval,we check if $f'(x) \ge 0$ for all $x$ in that interval.$(a)$ For $f(x) = 3x^2 - 2x

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$\left( -\infty, \frac{1}{3} \right]$ | $3x^2 - 2x + 1$

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(A) To determine if a function $f(x)$ is increasing on an interval,we check if $f'(x) \ge 0$ for all $x$ in that interval.$(a)$ For $f(x) = 3x^2 - 2x + 1$,$f'(x) = 6x - 2$. Setting $f'(x) \ge 0$ gives $6x \ge 2$,so $x \ge \frac{1}{3}$. The function is increasing on $[\frac{1}{3}, \infty)$,not $(-\infty, \frac{1}{3}]$. Thus,option $(a)$ is incorrectly matched.$(b)$ For $f(x) = x^3 + 6x^2 + 6$,$f'(x) = 3x^2 + 12x = 3x(x + 4)$. $f'(x) \ge 0$ when $x \in (-\infty, -4] \cup [0, \infty)$. Thus,it is increasing on $(-\infty, -4]$.$(c)$ For $f(x) = x^3 - 3x^2 + 3x + 3$,$f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2$. Since $3(x - 1)^2 \ge 0$ for all $x \in \mathbb{R}$,the function is increasing on $(-\infty, \infty)$.$(d)$ For $f(x) = 2x^3 - 3x^2 - 12x + 6$,$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$. $f'(x) \ge 0$ when $x \in (-\infty, -1] \cup [2, \infty)$. Thus,it is increasing on $[2, \infty)$.

$A$ function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval | Function

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$A$ function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?Interval | Function