True or False concept questions

[Taliaferro]

i) If f''(a) = 0, f is neither concave up nor concave down around x=a.
True

ii) If f is continuous on ( a,b ) and c is a local maximizer then f'(c)=0
True

iii) If f is continuous but not necessarily differentiable on [0,1], then the absolute maximum and the absolute minimum of f exist.
False

iv) If f is differentiable on ( a,b ) then it is also continuous on ( a,b ) and the absolute maximum and absolute minimum exist.
True

v) If x=a corresponds to an inflection point of f, then f''(a)=0 and f''(x) changes sign around x=a
False

Underneath are the answers I came up with. If I got any of them wrong could you please help me understand why?

Number iii) was a guess for me, I'm not quite sure what clues "continuous but not necessarily differentiable" should give me about this problem (but something about it seems suspiciously false).

For v) if I understand correctly inflection points are where the functions shape change from convex/concave up/down, correct? But that doesnt necessarily mean that the second derivative changes sign, right? Thats why I chose False for that one.

Thank you!

 

[HallsofIvy]

i) If f''(a) = 0, f is neither concave up nor concave down around x=a.
True

ii) If f is continuous on ( a,b ) and c is a local maximizer then f'(c)=0
True

iii) If f is continuous but not necessarily differentiable on [0,1], then the absolute maximum and the absolute minimum of f exist.
False

This is incorrect. A function continuous on a closed and bounded interval has both maximum and minimum values on that interval. Differentiability is not relevant.

iv) If f is differentiable on ( a,b ) then it is also continuous on ( a,b ) and the absolute maximum and absolute minimum exist.
True

It is certainly true that if a function is differentiable on (a, b) it is continuous on that interval. But it does not follow that it takes on maximum or minimum values on that interval. f(x)= x on (0, 1) is an example.

v) If x=a corresponds to an inflection point of f, then f''(a)=0 and f''(x) changes sign around x=a
False

See below

Underneath are the answers I came up with. If I got any of them wrong could you please help me understand why?

Number iii) was a guess for me, I'm not quite sure what clues "continuous but not necessarily differentiable" should give me about this problem (but something about it seems suspiciously false).

For v) if I understand correctly inflection points are where the functions shape change from convex/concave up/down, correct? But that doesnt necessarily mean that the second derivative changes sign, right? Thats why I chose False for that one.

If a function is concave upward then its second derivative is positive, right? And if it is concave downward its second derivative is negative? So how could it change from "concave upward" to "concave downward" without the second derivative changing sign?

Thank you!

 

[Taliaferro]

This is incorrect. A function continuous on a closed and bounded interval has both maximum and minimum values on that interval. Differentiability is not relevant.

Continuity vs Differentiability is definitely going to be on my final tomorrow. You just cleared a lot of things up for me. I see now that if a continuous function is defined at the endpoints then a max and a min value is on the interval.

It is certainly true that if a function is differentiable on (a, b) it is continuous on that interval. But it does not follow that it takes on maximum or minimum values on that interval. f(x)= x on (0, 1) is an example.

It seems like this is an extension of reasoning from what I just learned from my previous mistake. Since its an open interval (0, 1) vs [0, 1] then the function does not necessarily have max or min values. Right?
(Also wasnt clear on whether differentiablility on an interval meant the function was continuous, good to know)

See below


If a function is concave upward then its second derivative is positive, right? And if it is concave downward its second derivative is negative? So how could it change from "concave upward" to "concave downward" without the second derivative changing sign?

Yes, I was looking directly at this picture (see attached) of the graph and I completely missed the point until you clarified for me just now.

Sincerely, THANK YOU for your time and help, HallsofIvy!