derivatives and continuity

[pokitoki]

Given: Let f be a function such that lim (as h approaches 0) (f(2+h)-f(2))/h=5. In other words, f ' (2)=5.
Question: Which of the following must be true:
I. f is continuous at x=2
II. f is differentiable at x=2
III. the derivative of f is continuous at x=2.
Answer choices:
A. I only B. II only C. I and II only D. I and III only E. II and III only

My dilemma: I know the answer is C. (I and II only), but I'm trying to find an example of a function and its graph that fits this description (I'd like a graph that illustrates that III does not have to be true). Any thoughts?

 

[Deleted member 4993]

Given: Let f be a function such that lim (as h approaches 0) (f(2+h)-f(2))/h=5. In other words, f ' (2)=5.
Question: Which of the following must be true:
I. f is continuous at x=2
II. f is differentiable at x=2
III. the derivative of f is continuous at x=2.
Answer choices:
A. I only B. II only C. I and II only D. I and III only E. II and III only

My dilemma: I know the answer is C. (I and II only), but I'm trying to find an example of a function and its graph that fits this description (I'd like a graph that illustrates that III does not have to be true). Any thoughts?

Lots of thoughts - but what are yours?

Start with definition of continuity and existence - what kind of function can exist but not be continuous? how about '|x|' type of functions?

 

[mmm4444bot]

… what kind of function can exist but its first derivative will not be continuous? how about '|x|' type of functions?

I tried to clarify what I think Subhotosh means to say by inserting the words in red above; I hope that I clarified correctly. (Lately, I seem to be misinterpreting a lot of what is posted at this site.)

 

[Deleted member 4993]

Re:

I tried to clarify what I think Subhotosh means to say by inserting the words in red above; I hope that I clarified correctly. (Lately, I seem to be misinterpreting a lot of what is posted at this site.)

Yes indeed you clarified my cloudy writing correctly ....

 

[pokitoki]

The function f does not merely need to exist, it must be DIFFERENTIABLE at x=2. (|x| doesn't fit the bill) I need f ' (f prime) to be discontinuous at x=2. So in other words, f ' has a value at x=2, but its graph (f prime's graph) is discontinuous at x=2. Can you picture (graph) this? By the way, this question came off of an old AP Calc exam.

 

[Deleted member 4993]

The function f does not merely need to exist, it must be DIFFERENTIABLE at x=2. (|x| doesn't fit the bill)

Ofcourse it does.

You need to think

f'(x) = |x-2| + 5

f"(x) does not exist at x = 2 - but f'(x) exists.

Now find f(x).

f(x) = x[sup:3gsfci0r]2[/sup:3gsfci0r]/2 + 3x for x ? 2

f(x) = 7x - x[sup:3gsfci0r]2[/sup:3gsfci0r]/2 - 4 for x ? 2

plot it in your graphical calculator


I need f ' (f prime) to be discontinuous at x=2. So in other words, f ' has a value at x=2, but its graph (f prime's graph) is discontinuous at x=2. Can you picture (graph) this? By the way, this question came off of an old AP Calc exam.

 

[mmm4444bot]

… |x| doesn't fit the bill …

I'm thinking that you did not read closely enough. The suggestion is not |x|. The suggestion is |x| type functions.

The modifier "type" has meaning! It's a part of what we English speakers call a compound adjective.