Let $f, g: [-1, 2] \rightarrow \mathbb{R}$ be continuous functions which are twice differentiable on the interval $(-1, 2)$. Let the values of $f$ and $g$ at the points $-1$, $0$ and $2$ be as given in the following table: | | $x = -1$ | $x = 0$ | $x = 2$ | | --- | --- | --- | --- | | $f(x)$ | 3 | 6 | 0 | | $g(x)$ | 0 | 1 | -1 | In each of the intervals $(-1, 0)$ and $(0, 2)$ the function $(f - 3g)''$ never vanishes. Then the correct statement(s) is(are)

Let f, g: [-1, 2] → R be continuous functions which are twice differentiable on the interval (-1, 2). Let the values of f and g at the points -1, 0 and 2 be as given in the following table: | | x = -1 | x = 0 | x = 2 | | --- | --- | --- | --- | | f(x) | 3 | 6 | 0 | | g(x) | 0 | 1 | -1 | In each of the intervals (-1, 0) and (0, 2) the function (f - 3g)'' never vanishes. Then the correct statement(s) is(are) $f(-1)=3,\quad f(0)=6,\quad f(2)=0$ $g(-1)=0,\quad g(0)=1,\quad g(2)=-1$ $h=f-3g$ $\text{Find the number of solutions of } h'(x)=0 \text{ in } (-1,0) \text{ and } (0,2)$ $\text{Given: } h''(x)\ne 0 \text{ on } (-1,0) \text{ and } (0,2)$ $h=f-3g, \quad h'=f'-3g'$ $1.\ \text{Rolle on } [-1,0] \text{ and } [0,2]$ $2.\ h''\ne0 \Rightarrow \text{at most one zero each}$ $h(-1)=f(-1)-3g(-1)=3-0=3$ $h(0)=f(0)-3g(0)=6-3=3$ $h(2)=f(2)-3g(2)=0+3=3$ $h(-1)=h(0)=3 \Rightarrow \exists\, c_1\in(-1,0):\ h'(c_1)=0$ $h(0)=h(2)=3 \Rightarrow \exists\, c_2\in(0,2):\ h'(c_2)=0$ $h''\ne0 \Rightarrow h' \text{ strictly monotone on each interval}$ $\text{two zeros of } h' \Rightarrow \text{some } h''=0, \text{ contradiction}$ $\text{exactly one zero in } (-1,0), \text{ exactly one in } (0,2)$ $\Rightarrow \text{total two in } (-1,0)\cup(0,2)$ $\text{(2) one in } (-1,0): \text{TRUE}; \quad \text{(3) one in } (0,2): \text{TRUE}$ $\text{(1) three total, (4) two each}: \text{FALSE}$ $\boxed{\text{Answer: (2) and (3)}}$