Twice Differentiable Function: Two questions, please help! Thanks!!!?

1.) Since f'(3) = 0 there is a critical point at x = 3, and since f''(3) < 0 there

must be a local maximum at x = 3, so option (A) is correct.

(Note that since f(x) is twice differentiable on [0, 7] it must necessarily be

continuous on [0, 7], thus the conditions for the MVT are satisfied.)

2.) There may be other critical points to check, so (D) is the correct option.

Edit: Yes, there must be at least one turning point on 3 < x < 7, but it could

be that f(x) then rises to a value greater than 6 for x < 7 and then drops back

to f(7) = 6. I interpreted "twice differentiable" to mean at least twice differentiable,

and that there could be other values of x such that f'(x) = 0. As you say, it would

be paradoxical, (inconsistent might be a better word choice), if we were also

told that the only critical point was x = 3. The only way that would work is if

there was an inflection point at x = 3, but since we are given that f''(3) = -2 there

can't be an inflection point at x = 3.

Edit #2: Interesting. I'm trying to think of a function that is only twice

differentiable. The function f(x) = x^2 has f'(x) = 2x, f''(x) = 2 and higher

derivatives all 0, so f(x) in this case is infinitely differentiable.

We're probably over-thinking the question, but just out of curiosity, consider

the function f(x) = {(x - 3)^4 for x < 3 and (x - 3)^3 for x >= 3} then

f''(x) = {12*(x - 3)^2 for x < 3 and 6*(x - 3) for x > 3}, and so

lim(x->3-)(f''(x)) = lim(x->3+)(f''(x)) = 0, and so f(x) is twice differentiable

for all x including x = 3. However, f'''(x) = {24*(x - 3) for x < 3 and 6 for x > 3},

so lim(x->3-)(f'''(x)) = 0 and lim(x->3+)(f'''(x)) = 6, so lim(x->3)(f'''(x)) does not

exist, and hence f(x) is not thrice differentiable at x = 3. So we can create

a function that is only twice differentiable at a certain point, and thus is only

twice differentiable on any interval containing this point. This function doesn't

relate precisely to the question at hand, but it does open the door to the

possibility of creating a function that does.

My sense, however, is that what you are thinking is that the function only has two

critical points, such as a degree 3 polynomial. In that case then, yes, C would

be the correct answer. So in summary, the wording of the question does present

some ambiguity, in the sense that you could make the case for C or D.