The image shows a mathematical problem asking about a continuous function $f$ on the interval $[0, 2]$ with given values for $f(0) = 1$, $f(1) = 3$, and $f(2) = 2$. The question asks how many solutions the equation $f(x) = \frac{5}{2}$ must have on the interval $[0, 2]$, using the Intermediate Value Theorem (IVT).

Solution:

The Intermediate Value Theorem states that if a function $f$ is continuous on a closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f(c) = N$.

In this case:

• $f(0) = 1$
• $f(1) = 3$
• $f(2) = 2$

We are looking for solutions to $f(x) = \frac{5}{2} = 2.5$.

1. On the interval $[0, 1]$:

   • $f(0) = 1$
   • $f(1) = 3$ Since $2.5$ is between $f(0) = 1$ and $f(1) = 3$, by the IVT, there must be at least one solution for $f(x) = 2.5$ in the interval $(0, 1)$.

2. On the interval $[1, 2]$:

   • $f(1) = 3$
   • $f(2) = 2$ Again, since $2.5$ is between $f(1) = 3$ and $f(2) = 2$, there must be at least one solution for $f(x) = 2.5$ in the interval $(1, 2)$.

Thus, the equation $f(x) = 2.5$ must have two solutions: one in $(0, 1)$ and one in $(1, 2)$.

Answer: (B) 2

Let me know if you need further clarification or a breakdown of the concepts.

Related Questions:

1. What is the Intermediate Value Theorem, and how does it apply to this problem?
2. Can the Intermediate Value Theorem guarantee the exact number of solutions?
3. What would happen if the function were not continuous on the interval $[0, 2]$?
4. How would the problem change if $f(1)$ were less than 2.5?
5. Can you generalize this result to other functions and intervals?

Tip: Always check for the continuity of the function before applying the Intermediate Value Theorem.