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

To solve this problem using the **Intermediate Value Theorem (IVT)**, we need to check where the function \( h(x) \) crosses the value \( h(x) = -20 \).

### Intermediate Value Theorem (IVT):
The IVT states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and \( f(a) \) and \( f(b) \) have different signs (i.e., \( f(a) < c < f(b) \) or \( f(a) > c > f(b) \)), then there must exist some \( x \) in \((a, b)\) such that \( f(x) = c \).

### Given table of values for \( h(x) \):

\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 1 & 2 & 3 & 4 & 5 \\
\hline
h(x) & -98 & -32 & 15 & -27 & 2 \\
\hline
\end{array}
\]

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.

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

Solve a problem using the Intermediate Value Theorem to determine how many times h(x) = −20 for a continuous function with given values. Learn how IVT ensures the existence of roots within intervals where the function changes sign. Suitable for Grades 11-12 calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation