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

To solve for the values of \( k \) such that the equation \( f(x) = 2 \) has at least two solutions in the interval \([-1, 1]\), let's analyze the problem step by step.

### Step 1: Understanding the problem
We are given that the function \( f(x) \) is continuous on the interval \([-1, 1]\), and the values of \( f(x) \) are provided at three points:
- \( f(-1) = 3 \)
- \( f(0) = k \)
- \( f(1) = 5 \)

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

### Step 2: Applying the Intermediate Value Theorem
Since \( f(x) \) is continuous, we can apply the **Intermediate Value Theorem** (IVT). This theorem states that if a function is continuous on a closed interval \([a, b]\) and takes values \( f(a) \) and \( f(b) \) at the endpoints, then for any value \( y \) between \( f(a) \) and \( f(b) \), there exists at least one \( c \in (a, b) \) such that \( f(c) = y \).

We are looking for the values of \( k \) where \( f(x) = 2 \) has at least two solutions. Let's examine the intervals \([-1, 0]\) and \([0, 1]\) separately.

### Step 3: Finding solutions in the interval \([-1, 0]\)
- On the interval \([-1, 0]\), we know that \( f(-1) = 3 \) and \( f(0) = k \).
- For the equation \( f(x) = 2 \) to have a solution in this interval, \( 2 \) must lie between \( f(-1) = 3 \) and \( f(0) = k \).
- This gives the condition:
  \[
  3 \geq 2 \geq k
  \]
  or equivalently, \( k \leq 2 \).

### Step 4: Finding solutions in the interval \([0, 1]\)
- On the interval \([0, 1]\), we know that \( f(0) = k \) and \( f(1) = 5 \).
- For the equation \( f(x) = 2 \) to have a solution in this interval, \( 2 \) must lie between \( f(0) = k \) and \( f(1) = 5 \).
- This gives the condition:
  \[
  k \leq 2 \leq 5
  \]
  or equivalently, \( k \leq 2 \).

### Step 5: Combining the conditions
From both intervals, we obtained the condition \( k \leq 2 \). This ensures that there is at least one solution in each interval, leading to **two solutions** in total if \( k < 2 \).

### Final Answer:
For the equation \( f(x) = 2 \) to have at least two solutions, \( k \) must satisfy:
\[
k < 2
\]
If \( k = 2 \), there would be exactly one solution at \( x = 0 \).

---

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How does the Intermediate Value Theorem apply to functions that are not continuous?
2. What happens to the number of solutions if \( k > 2 \)?
3. Can you explain the difference between "at least two solutions" and "exactly two solutions" in this context?
4. How would the solution change if the function was not continuous on \([-1, 1]\)?
5. What other types of problems can be solved using the Intermediate Value Theorem?

**Tip:** The Intermediate Value Theorem is a powerful tool for proving the existence of solutions but does not guarantee how many solutions exist. Always check boundary conditions carefully.

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

Determine the values of k for which the equation f(x) = 2 has at least two solutions in the interval [-1, 1]. The function f(x) is continuous, and the Intermediate Value Theorem is applied to analyze the solutions based on the given values of f(x) at specific points. Suitable for high school students studying advanced calculus.

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