Math Problem Statement
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
Solution
We are given a function that is continuous on the closed interval , with the following values from the table:
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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Intermediate Value Theorem
Equations
Formulas
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Theorems
Intermediate Value Theorem
Suitable Grade Level
Grades 11-12
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