We are tasked with determining the validity of the statement:

"$f(x) = 1$ has at least 2 solutions in the interval $2 \leq x \leq 7$."

Given Data:

• The function $f(x)$ is continuous.
• Values of $f(x)$:
  • $f(2) = 3$
  • $f(4) = -1$
  • $f(7) = 5$

Analysis:

We will use the Intermediate Value Theorem (IVT), which states:
If a function $f(x)$ is continuous on a closed interval $[a, b]$, and $y$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that $f(c) = y$.

Step 1: Identify where $f(x) = 1$ occurs.

1. On the interval $[2, 4]$:

   • $f(2) = 3$ and $f(4) = -1$.
   • Since $f(x)$ is continuous and $1$ lies between $3$ and $-1$, the IVT guarantees at least one solution to $f(x) = 1$ in $(2, 4)$.

2. On the interval $[4, 7]$:

   • $f(4) = -1$ and $f(7) = 5$.
   • Similarly, $f(x) = 1$ lies between $-1$ and $5$. By the IVT, there is at least one solution in $(4, 7)$.

Step 2: Determine if there are at least two distinct solutions.

• One solution exists in $(2, 4)$, and another in $(4, 7)$.
• These solutions are distinct since they occur in disjoint intervals.

Conclusion:

The statement "$f(x) = 1$ has at least 2 solutions in $2 \leq x \leq 7$" must be true.

Correct answer: The statement must be true.

Would you like a deeper explanation of the Intermediate Value Theorem or how continuity is applied here?
Here are 5 related questions to expand understanding:

1. What is the Intermediate Value Theorem, and how is it applied in real-world scenarios?
2. Can discontinuous functions satisfy the conditions of the IVT?
3. What would happen if $f(4) = 1$? How would the result change?
4. How can we generalize this problem for finding $f(x) = k$ for any $k$?
5. How does continuity ensure the existence of solutions within an interval?

Tip: Always confirm if a function is continuous before applying the Intermediate Value Theorem.