Question:

Step 1: Note the given conditions. $f: \mathbb{R} \to [-3, 3]$, $f$ is twice differentiable, $f'(0) = f(1) = f(3) = 2$. Step 2: Verify option (a). Since $f$ is differentiable, it is continuous. On any interval where $f'(x) > 0$ or $f'(x) < 0$, $f$ is monotonic. Since $f$ is differentiable and not constant (it takes value 2 at $x=1$ and $x=3$), there exist intervals where it is monotonic. Option (a) is correct. Step 3: Verify option (b). $f'(0) = 2$. Since $f: \mathbb{R} \to [-3,3]$, $f$ is bounded. By the mean value theorem applied on $[-3, 0]$: there exists $c \in (-3, 0)$ such that $f'(c) = \dfrac{f(0) - f(-3)}{0-(-3)} = \dfrac{f(0)-f(-3)}{3}$. Since $|f(0)| \leq 3$ and $|f(-3)| \leq 3$, $|f(0)-f(-3)| \leq 6$, so $|f'(c)| \leq 2$. Thus $f'(c) \leq 2$ for some $c \in (-3,0)$. Option (b) is correct. Step 4: Verify option (c). Apply MVT on $[1,3]$: $\exists\, c_1 \in (1,3)$ such that $f'(c_1) = \dfrac{f(3)-f(1)}{3-1} = \dfrac{2-2}{2} = 0$. Now apply MVT for $f'$ on $[c_1, \text{ some point}]$ or use the fact that $f'(0)=2$ and $f'(c_1)=0$. Apply MVT on $[0, c_1]$: $\exists\, c_2 \in (0, c_1)$ such that $f''(c_2) = \dfrac{f'(c_1)-f'(0)}{c_1-0} = \dfrac{0-2}{c_1} = \dfrac{-2}{c_1}$. Since $c_1 \in (1,3)$, $\dfrac{-2}{c_1} \in (-2, -\frac{2}{3})$. So $f''(c_2) \geq \dfrac{-2}{c_1} > -2$. Also $f''(c_2) = \dfrac{-2}{c_1} \geq \dfrac{-2}{3} > \dfrac{-1}{3}$... Actually $\dfrac{-2}{c_1}$ with $c_1 \in (1,3)$ gives $f''(c_2) \in (-2, -2/3)$, so $f''(c_2) \geq -2$. For option (c): $f''(x) \geq -1/3$ for some $c$. Since $f''$ is continuous and $f''(c_2) < 0$, but we need to check if $f''$ takes values $\geq -1/3$. Since $f'(0)=2$ and $f'$ must return (by boundedness of $f$), $f''$ must be positive somewhere, so $f''(c) \geq -1/3$ (in fact $f''(c) > 0$) for some $c$. Option (c) is correct. Step 5: Verify option (d). From Step 4, $f''(c_2) = -2/c_1 > -2$ (since $c_1 < 3$... wait $c_1 \in (1,3)$ so $-2/c_1 > -2$). So $f''(c_2) \geq -2$ is satisfied. Option (d) is correct.