Suppose the function f (x) is continuous on the closed interval [−1, 1] and has values that are
given in the table below. For what values of k will the equation f (x) = 2 have at least two
solutions in the interval [−1, 1] ? table: x | -1 | 0 | 1 f(x) | 3 | k | 5

We are given a function \( f(x) \) that is continuous on the closed interval \([-1, 1]\), with the following values from the table:

\[
\begin{array}{c|c|c|c}
x & -1 & 0 & 1 \\
\hline
f(x) & 3 & k & 5 \\
\end{array}
\]

We are tasked with determining the values of \( k \) such that the equation \( f(x) = 2 \) has at least two solutions in the interval \([-1, 1]\).

### Step 1: Intermediate Value Theorem (IVT)
Since the function \( f(x) \) is continuous on \([-1, 1]\), the **Intermediate Value Theorem** guarantees that if \( f(x) \) takes values above and below 2 at different points, then there exists at least one point where \( f(x) = 2 \).

For the equation \( f(x) = 2 \) to have **two solutions**, the function must cross \( f(x) = 2 \) at least twice within the interval.

### Step 2: Analyze Behavior at Given Points

From the table:

- At \( x = -1 \), \( f(-1) = 3 \),
- At \( x = 0 \), \( f(0) = k \),
- At \( x = 1 \), \( f(1) = 5 \).

Thus, the function takes the following values at the endpoints:
\[
f(-1) = 3 \quad \text{and} \quad f(1) = 5.
\]

We want \( f(x) = 2 \) to have at least two solutions. Since \( f(-1) = 3 \) and \( f(1) = 5 \), both values are **greater than 2**. Therefore, for \( f(x) = 2 \) to have two solutions, \( k \) must be **less than 2** so that the function crosses \( f(x) = 2 \) twice—once between \( x = -1 \) and \( x = 0 \), and once between \( x = 0 \) and \( x = 1 \).

### Step 3: Determine the Range for \( k \)

For \( f(x) = 2 \) to have at least two solutions:
- The function must drop below 2 at \( x = 0 \), which requires \( f(0) = k < 2 \).
- Additionally, \( k \) should not be too low to prevent crossing the value \( f(x) = 2 \) between \( x = -1 \) and \( x = 0 \), meaning that \( k > -\infty \).

### Conclusion:

The value of \( k \) must satisfy \( k < 2 \) for the equation \( f(x) = 2 \) to have at least two solutions in the interval \([-1, 1]\).

---

Do you want more details or have any questions?

Here are five related questions to expand on this topic:

1. How does the Intermediate Value Theorem help determine the existence of solutions?
2. What happens if \( k = 2 \)? How many solutions would there be in that case?
3. Can there be more than two solutions for \( f(x) = 2 \) in this case?
4. How does the behavior of \( f(x) \) between \( x = -1 \) and \( x = 1 \) affect the number of solutions?
5. What would happen if \( k > 5 \)? Would there be any solutions for \( f(x) = 2 \)?

**Tip:** Always visualize the graph of the function when applying the Intermediate Value Theorem to better understand the behavior of the function over an interval.

Suppose the function f(x) is continuous on the closed interval [-1, 1] and has values that are given in the table below. For what values of k will the equation f(x) = 2 have at least two solutions in the interval [-1, 1]? Table: x | -1 | 0 | 1, f(x) | 3 | k | 5.

Explore how to determine the values of k such that the equation f(x) = 2 has at least two solutions in the interval [-1, 1]. This problem uses the Intermediate Value Theorem and analyzes the continuity of a function. Suitable for advanced high school students in Grades 11-12.

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