Math Problem Statement
Selected values of a continuous function h are shown in the table.
x 1 2 3 4 5 h(x) −98 −32 15 −27 2
According to the IVT, which of the following is the fewest possible number of times h(x) = −20? Three Two One Zero
Solution
To solve this problem using the Intermediate Value Theorem (IVT), we need to check where the function crosses the value .
Intermediate Value Theorem (IVT):
The IVT states that if a function is continuous on a closed interval , and and have different signs (i.e., or ), then there must exist some in such that .
Given table of values for :
\hline x & 1 & 2 & 3 & 4 & 5 \\ \hline h(x) & -98 & -32 & 15 & -27 & 2 \\ \hline \end{array}$$ We want to find where $$h(x) = -20$$. Using the IVT, we need to check where $$h(x)$$ changes from a value less than $$-20$$ to a value greater than $$-20$$, or vice versa. ### Step-by-step analysis: 1. Between $$x = 1$$ and $$x = 2$$: - $$h(1) = -98$$ and $$h(2) = -32$$ - Both values are less than $$-20$$, so no crossing of $$h(x) = -20$$ here. 2. Between $$x = 2$$ and $$x = 3$$: - $$h(2) = -32$$ and $$h(3) = 15$$ - $$h(2) < -20$$ and $$h(3) > -20$$, so by the IVT, $$h(x) = -20$$ **at least once** between $$x = 2$$ and $$x = 3$$. 3. Between $$x = 3$$ and $$x = 4$$: - $$h(3) = 15$$ and $$h(4) = -27$$ - $$h(3) > -20$$ and $$h(4) < -20$$, so by the IVT, $$h(x) = -20$$ **at least once** between $$x = 3$$ and $$x = 4$$. 4. Between $$x = 4$$ and $$x = 5$$: - $$h(4) = -27$$ and $$h(5) = 2$$ - $$h(4) < -20$$ and $$h(5) > -20$$, so by the IVT, $$h(x) = -20$$ **at least once** between $$x = 4$$ and $$x = 5$$. ### Conclusion: - $$h(x) = -20$$ occurs at least **once between $$x = 2$$ and $$x = 3$$**, **once between $$x = 3$$ and $$x = 4$$**, and **once between $$x = 4$$ and $$x = 5$$**. Thus, the fewest possible number of times that $$h(x) = -20$$ is **three**. ### Answer: **Three** --- Would you like further details on the Intermediate Value Theorem, or any additional questions? Here are 5 related questions: 1. What are the conditions for the Intermediate Value Theorem to apply? 2. Can the IVT be applied to discontinuous functions? 3. What would happen if there were only two intervals where $$h(x) = -20$$? 4. How can we estimate the location of the $$x$$-values where $$h(x) = -20$$? 5. Is it possible for $$h(x)$$ to equal $$-20$$ more than three times in this scenario? **Tip:** Always check the continuity of a function before applying the Intermediate Value Theorem.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Intermediate Value Theorem
Continuity of Functions
Formulas
None (The Intermediate Value Theorem applies based on function properties rather than a specific formula)
Theorems
Intermediate Value Theorem
Suitable Grade Level
Grades 11-12 (Advanced High School Calculus)
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