Math Problem Statement
The function f is continuous on the closed interval [-2, 2]. Several values of the function f are given in the table below. The equation f(x) = 1 must have at least two solutions in the interval [-2, 2] if c = ?
Solution
We are given a continuous function on the closed interval , and we need to determine the value of such that the equation has at least two solutions in the interval .
From the table:
x & -2 & 0 & 2 \\ \hline f(x) & 3 & c & 2 \\ \end{array}$$ ### Step 1: Understand the Intermediate Value Theorem (IVT) The function $$f$$ is continuous on the interval $$[-2, 2]$$, and the IVT guarantees that if $$f(a) \neq f(b)$$ and the value $$N$$ lies between $$f(a)$$ and $$f(b)$$, then there exists a point $$c \in (a, b)$$ such that $$f(c) = N$$. ### Step 2: Analyze $$f(x) = 1$$ We need to find the value of $$c$$ such that $$f(x) = 1$$ has at least two solutions. - At $$x = -2$$, we know $$f(-2) = 3$$. - At $$x = 2$$, we know $$f(2) = 2$$. - At $$x = 0$$, we have $$f(0) = c$$. We want $$f(x) = 1$$ to have at least two solutions. The value of $$c$$ needs to be such that the graph of the function crosses $$f(x) = 1$$ at least twice. ### Step 3: Apply IVT 1. Consider the interval $$[-2, 0]$$. - $$f(-2) = 3$$ and $$f(0) = c$$. - For $$f(x) = 1$$ to have a solution in this interval, we need $$c \leq 1$$, because $$f(-2) = 3 > 1$$ and the function needs to decrease to at least 1 to ensure a solution in this interval. 2. Consider the interval $$[0, 2]$$. - $$f(0) = c$$ and $$f(2) = 2$$. - For $$f(x) = 1$$ to have a solution in this interval, we need $$c \geq 1$$, because $$f(2) = 2 > 1$$. Thus, to have at least two solutions for $$f(x) = 1$$, we require $$c = 1$$ to satisfy both conditions. ### Conclusion: The value of $$c$$ must be $$\boxed{1}$$ for $$f(x) = 1$$ to have at least two solutions in the interval $$[-2, 2]$$. --- Would you like further clarification or details on this solution? Here are some related questions to expand on this topic: 1. What would happen if $$c > 1$$ or $$c < 1$$? 2. Can the Intermediate Value Theorem guarantee more than two solutions for a continuous function? 3. How does continuity of $$f(x)$$ play a role in this problem? 4. Could this problem be solved without using the Intermediate Value Theorem? 5. What if $$f(x)$$ were discontinuous—would the problem change? **Tip:** Always check the behavior of the function at the boundary points when applying the Intermediate Value Theorem.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity of Functions
Intermediate Value Theorem
Solving Equations
Formulas
-
Theorems
Intermediate Value Theorem
Suitable Grade Level
Grades 11-12
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