The problem states that the function $f(x)$ is continuous for all real numbers $x$, with the following given conditions:

• $f(3) = -2$
• $f(7) = 1$
• $f(x) = 0$ for exactly one value of $x$.

We are tasked with finding which value of $x$ is possible for $f(x) = 0$, given the options:

• a. $-1$
• b. $0$
• c. $2$
• d. $4$
• e. $8$

Solution:

Since $f(x)$ is continuous, by the Intermediate Value Theorem (IVT), if $f(a)$ and $f(b)$ have opposite signs, there must be at least one value $c$ between $a$ and $b$ such that $f(c) = 0$.

From the given information:

• $f(3) = -2$ (negative)
• $f(7) = 1$ (positive)

The Intermediate Value Theorem guarantees that there is at least one $x \in (3, 7)$ such that $f(x) = 0$, because the function changes sign from negative to positive in this interval.

Now, we check the options:

• The only value between 3 and 7 is $x = 4$, which is listed as option d.

Thus, the correct answer is d. 4.

Would you like more details or have any questions?

Here are 5 related questions to further explore this concept:

1. How does the Intermediate Value Theorem apply to continuous functions?
2. Can a function have more than one root between two points where the function changes sign?
3. How would the problem change if $f(3) = 0$ instead of $f(3) = -2$?
4. What happens if a function is not continuous on a given interval?
5. What are some examples of discontinuous functions where the IVT doesn't apply?

Tip: The Intermediate Value Theorem only guarantees the existence of a root, but it doesn't tell you how to find it exactly. Methods like bisection are used to approximate roots numerically.