Which statement "must be false"? Given a function $f$ continuous on $[-4, 1]$ with its maximum at $(-3, 5)$ and its minimum at $(1/2, -6)$, is it not correct to say that both statements (B) and (D) must be false?
(A) The graph of $f$ crosses both axes.
(B) $f$ is always decreasing on $[-4, 1]$.
(C) $f(-2)=0$,
(D) $f(-1)=6$,
(E) $f(0)=2$.
If the maximum (on this interval at least) is $(-3, 5)$, then $(-1, 6)$ cannot be a point on the graph (D). That the function is "always decreasing" (B) on the interval seems to contradict both the stated maximum and stated minimum. The other options all seem either certainly true or possible. Am I missing something?

In my opinion, the conditions on $f$ aren't well-posed. I'll put the offending words in bold: 

Given a function $f$ continuous on $[-4, 1]$ with its maximum at
  (-3, 5) and its minimum at (1/2, -6)...

I feel that the phrasing mistakenly assumes that $f$ has a unique maximum and a unique minimum, which (obviously) doesn't follow from the assumption that $f$ is continuous on $[-4,1]$. The question should read:

Given a function $f$ continuous on $[-4, 1]$ with a maximum at
  (-3, 5) and a minimum at (1/2, -6)...

When phrased in this way, it becomes more clear that $f(x)$ may equal $5$ for all $x \in [-4,-3]$ and may equal $-6$ for all $x \in [1/2,1]$.
There's a further ambiguity: does decreasing mean $x < x' \rightarrow f(x) < f(x'),$ or does it mean the weaker $x \leq x' \rightarrow f(x) \leq f(x')$? If it means the weaker of the two conditions, then (B) needn't be false. So (D) is definitely "more false" than (B). But still, the question simply isn't clear.
I have another issue with this question (not your question, but rather, the question you've been asked.) Namely that it indulges in a common yet unfortunate convention. In particular, notice that the question never tells the reader that the codomain of $f$ is meant to be $\mathbb{R}$. And sure, you can infer it from the context; nonetheless I think its sloppy not to tell the reader these things explicitly.
