1. **Problem statement:** We are asked to determine the truth value of several statements about a function $f$ given only its variation table: $$\begin{array}{c|ccccc} x & -4 & -2 & 0 & 4 & 6 \\ f(x) & 1 & 0 & 3 & -3 & -1 \\ \end{array}$$ 2. **Recall:** The variation table shows values of $f(x)$ at specific points and the behavior (increasing or decreasing) between these points. 3. **Analyze each statement:** **a. $f(-3) = 4$** - $-3$ is between $-4$ and $-2$. - $f$ decreases from $1$ at $-4$ to $0$ at $-2$. - So $f(-3)$ is between $1$ and $0$, definitely not $4$. - **Answer:** False. **b. $f(1) > f(3)$** - $1$ and $3$ lie between $0$ and $4$. - $f$ decreases from $3$ at $0$ to $-3$ at $4$. - So $f(1) > f(3)$ since function is decreasing. - **Answer:** True. **c. $f(-1)$ is positive** - $-1$ is between $-2$ and $0$. - $f$ increases from $0$ at $-2$ to $3$ at $0$. - So $f(-1)$ is between $0$ and $3$, thus positive. - **Answer:** True. **d. $f(x) = 0$ has a single solution** - From table, $f(-2) = 0$. - Between $-4$ and $-2$, $f$ decreases from $1$ to $0$. - Between $-2$ and $0$, $f$ increases from $0$ to $3$. - No other zero values indicated. - **Answer:** True. **e. $f(1) > 3$** - $1$ is between $0$ and $4$ where $f$ decreases from $3$ to $-3$. - So $f(1) < 3$. - **Answer:** False. **f. $f(5)$ is negative** - $5$ is between $4$ and $6$. - $f$ increases from $-3$ at $4$ to $-1$ at $6$. - Both values negative, so $f(5)$ is negative. - **Answer:** True. **g. $f(-3) < f(-2)$** - $f(-2) = 0$. - $f$ decreases from $1$ at $-4$ to $0$ at $-2$. - So $f(-3)$ is between $1$ and $0$, thus $f(-3) > f(-2)$. - **Answer:** False. **h. If $x \in [0;6]$, then $f(x) \geq -3$** - $f(4) = -3$, $f(6) = -1$. - Between $0$ and $4$, $f$ decreases from $3$ to $-3$. - Between $4$ and $6$, $f$ increases from $-3$ to $-1$. - So $f(x) \geq -3$ on $[0;6]$. - **Answer:** True. **Final answers:** - a: False - b: True - c: True - d: True - e: False - f: True - g: False - h: True