1. **State the problem:** We are given values of a function $f$ at points $x=0,1,2,3$ and asked which statements about the Intermediate Value Theorem (IVT), Mean Value Theorem (MVT), and Extreme Value Theorem (EVT) must be true. 2. **Recall the theorems:** - IVT: If $f$ is continuous on $[a,b]$ and $N$ is between $f(a)$ and $f(b)$, then there exists $c \in (a,b)$ such that $f(c)=N$. - MVT: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$ - EVT: If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains a maximum and minimum value at some points in $[a,b]$. 3. **Analyze statement I:** - $f(0)=15$, $f(3)=9$, and $10$ lies between $9$ and $15$. - If $f$ is continuous on $[0,3]$, IVT guarantees $c \in (0,3)$ with $f(c)=10$. - Since the problem implies $f$ is a function with these values, continuity is assumed for IVT to apply. - Therefore, statement I must be true. 4. **Analyze statement II:** - Compute average rate of change: $$\frac{f(3)-f(0)}{3-0} = \frac{9-15}{3} = \frac{-6}{3} = -2.$$ - MVT requires $f$ to be continuous on $[0,3]$ and differentiable on $(0,3)$. - If these hold, there exists $c \in (0,3)$ with $f'(c) = -2$. - Since differentiability is not guaranteed by the table alone, statement II may or may not be true. 5. **Analyze statement III:** - EVT requires continuity on $[0,3]$. - If $f$ is continuous, it attains minimum and maximum on $[0,3]$. - The minimum value in the table is $9$ at $x=3$, so there exists $c$ with $f(c) \leq f(x)$ for all $x$ in $[0,3]$. - Thus, statement III must be true. **Final conclusion:** Statements I and III must be true; statement II depends on differentiability and is not guaranteed. **Answer:** I and III must be true.