If f is a continuous function such that f(1) = 3 and f(7) = −2, then its graph must intersect the x-axis. give a piecewise counterexample of this statement


The problem you're referring to involves the **Intermediate Value Theorem (IVT)**, which states that if a function \( f \) is continuous on a closed interval \([a, b]\) and takes different signs at \( f(a) \) and \( f(b) \), then there must be some \( c \in (a, b) \) such that \( f(c) = 0 \). However, you're asking for a counterexample that shows that **even with continuity**, a function may avoid intersecting the x-axis while still having \( f(1) = 3 \) and \( f(7) = -2 \).

### Piecewise Counterexample:

Define a **piecewise function** \( f(x) \) as follows:

\[
f(x) =
\begin{cases} 
3 & \text{for } x \in [1, 2) \\
-2 & \text{for } x \in [2, 7]
\end{cases}
\]

### Explanation:
- \( f(x) \) is **continuous** at every point within each interval.
- The function is not zero anywhere between \( x = 1 \) and \( x = 7 \).
- The graph remains **above the x-axis** (at 3) in the interval \( [1, 2) \), and **below the x-axis** (at -2) in the interval \( [2, 7] \).

Although \( f(1) = 3 \) and \( f(7) = -2 \), and \( f(x) \) is piecewise continuous, the function does not cross the x-axis at any point. This counterexample avoids satisfying the Intermediate Value Theorem because the continuity is only on certain intervals, and at \( x = 2 \), there is a **jump discontinuity**.

### Do you want more details or have any questions?

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### Related Questions:
1. What are the conditions required for the Intermediate Value Theorem to apply?
2. How can a piecewise function still be considered continuous?
3. Can the IVT be applied if a function is not continuous over the entire interval?
4. What types of discontinuities prevent a function from satisfying the IVT?
5. How can you adjust a piecewise function to ensure it satisfies the IVT?

#### Tip:
Always check the overall continuity of a function before applying the Intermediate Value Theorem; it must be continuous **everywhere** on the interval!

If f is a continuous function such that f(1) = 3 and f(7) = −2, then its graph must intersect the x-axis. Give a piecewise counterexample of this statement.

Explore a counterexample to the Intermediate Value Theorem with a continuous piecewise function where f(1) = 3 and f(7) = -2, but the function does not intersect the x-axis. Understand how discontinuities can prevent satisfying the IVT despite continuity on specific intervals.

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