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:

\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? --- ### 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!