If f(x) = begin{cases} 3x^2 + 12x - 1, & -1 l | vedclass

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(D) Given $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \ 37 - x, & 2 < x \le 3 \end{cases}$.$1$. Check for continuity at $x = 2$:$f(2) = 3(

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All of the above.

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(D) Given $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \ 37 - x, & 2 $1$. Check for continuity at $x = 2$:$f(2) = 3(2)^2 + 12(2) - 1 = 12 + 24 - 1 = 35$.$\lim_{x 	o 2^-} f(x) = 35$.$\lim_{x 	o 2^+} f(x) = 37 - 2 = 35$.Since $f(2) = \lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x)$,the function is continuous at $x = 2$ and thus on $[-1, 3]$.$2$. Check for monotonicity:For $x \in (-1, 2)$,$f'(x) = 6x + 12$. Since $x > -1$,$6x + 12 > 6(-1) + 12 = 6 > 0$. Thus,$f(x)$ is increasing on $[-1, 2]$.For $x \in (2, 3)$,$f'(x) = -1 $3$. Check for maximum:Since $f(x)$ increases on $[-1, 2]$ and decreases on $(2, 3]$,the maximum value occurs at $x = 2$,where $f(2) = 35$.Therefore,all statements are correct.

If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

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