The problem states:

Let $g$ be a continuous function on the closed interval $[0, 1]$, with $g(0) = 1$ and $g(1) = 0$. The task is to identify which of the following statements is NOT necessarily true:

(A) For all $a$ and $b$ in $[0, 1]$, if $a = b$, then $g(a) = g(b)$.
(B) There exists a number $h$ in $[0, 1]$ such that $g(h) = \frac{1}{2}$.
(C) There exists a number $h$ in $[0, 1]$ such that $g(h) = \frac{3}{2}$.
(D) For all $h$ in the open interval $(0, 1)$,
$\lim_{x \to h} g(x) = g(h).$

Analysis of the Options:

• (A): This statement is always true because $g$ is continuous, and by definition of continuity, if $a = b$, then $g(a) = g(b)$. Thus, this statement is true.

• (B): By the Intermediate Value Theorem, since $g(0) = 1$ and $g(1) = 0$, there must be some $h \in [0, 1]$ where $g(h) = \frac{1}{2}$. So, this statement is true.

• (C): This statement is false because $g(1) = 0$ and $g(0) = 1$, and the values of $g$ in the interval $[0, 1]$ are constrained between 0 and 1 due to the continuity of $g$. Therefore, there cannot be any $h$ in $[0, 1]$ such that $g(h) = \frac{3}{2}$, which is greater than 1.

• (D): This statement is a standard result from the definition of continuity, which holds for any continuous function on an interval. Thus, it is true.

Conclusion:

The statement that is NOT necessarily true is (C), since the function cannot take values greater than 1 in the interval $[0, 1]$.

Would you like further details or explanations?

Here are five related questions:

1. What is the Intermediate Value Theorem and how is it applied here?
2. Why does continuity imply that $g$ cannot take values larger than 1 in this context?
3. How does the limit definition in option (D) relate to the concept of continuity?
4. Can you provide an example of a continuous function that satisfies the conditions in the problem?
5. What would change if $g(0)$ and $g(1)$ were different values?

Tip: Continuity ensures that a function cannot "jump" in value within an interval—it must pass through all intermediate values.