basic of calculus

[bhuvaneshnick]

i have just started studying calculus day before .But few basic things messing up with me.
First

f(x) is continuous in the closed interval [a,b]

does that mean the that f(x) has non zero value for all values from a to b?

f'(x) is differentiable for every value in the the open interval(a,b)

does that mean for every value between a to b,the f'(x) would result in some value between a to b?
Thank you

 

[pka]

i have just started studying calculus day before .But few basic things messing up with me.
First

f(x) is continuous in the closed interval [a,b]

does that mean the that f(x) has non zero value for all values from a to b?

f'(x) is differentiable for every value in the the open interval(a,b)

does that mean for every value between a to b,the f'(x) would result in some value between a to b?

Consider: \(\displaystyle f(x)=x^3\) on \(\displaystyle [-1,2]\)

 

[Steven G]

i have just started studying calculus day before .But few basic things messing up with me.
First

f(x) is continuous in the closed interval [a,b]

does that mean the that f(x) has non zero value for all values from a to b?

f'(x) is differentiable for every value in the the open interval(a,b)

does that mean for every value between a to b,the f'(x) would result in some value between a to b?
Thank you

The interval [a,b] means that a<= x <= b.
When one says that f(x) is continuous on [a,b] it means that f(x) is continuous for a<= x <= b. The precise definition for f(x) to be continuous at x=c is
1) lim as x->c- = L
2) lim as x->c+ = L
and 3) f(c) = L

The fact that f(x) is continuous on [a,b] has NOTHING TO DO WITH THE VALUES OF f(x) on [a,b].

 

[pka]

The fact that f(x) is continuous on [a,b] has NOTHING TO DO WITH THE VALUES OF f(x) on [a,b].

I am totally mystified at your saying such.
Continuity on a closed interval tells me that the range of the function is a connected compact set. That is the image set is bounded and is itself a closed interval. From which we get the intermediate value theorem as well as the highpoint/lowpoint theorems. That seems to me to say a great deal about the values of \(\displaystyle f\).

 

[Ishuda]

I am totally mystified at your saying such.
Continuity on a closed interval tells me that the range of the function is a connected compact set. That is the image set is bounded and is itself a closed interval. From which we get the intermediate value theorem as well as the highpoint/lowpoint theorems. That seems to me to say a great deal about the values of \(\displaystyle f\).

And you are going to castigate Jomo for not adding that little word 'particular'? [but then I generally am]

 

[pka]

And you are going to castigate Jomo for not adding that little word 'particular'? [but then I generally am]

I have no idea what you are on about. There is nothing particular about the post to which I replied. It clearly is about continuity on a general interval.

 

[Steven G]

I am totally mystified at your saying such.
Continuity on a closed interval tells me that the range of the function is a connected compact set. That is the image set is bounded and is itself a closed interval. From which we get the intermediate value theorem as well as the highpoint/lowpoint theorems. That seems to me to say a great deal about the values of \(\displaystyle f\).

pka, I was answering this for a student who is studying calculus for one day. This student said that f(x) is never 0 on [a,b]. My response was mainly for that comment.
Having said that, you are 100% correct and I thank you for pointing this out.

 

[Ishuda]

I have no idea what you are on about. There is nothing particular about the post to which I replied. It clearly is about continuity on a general interval.

Oh, but you would agree that "The fact that f(x) is continuous on [a,b] has NOTHING TO DO WITH THE PARTICULAR VALUES OF f(x) on [a,b]." wouldn't you? That is, for example, given f(x) is continuous on [a, b] doesn't say anything about whether f(x) takes on the value of 2 in [a,b]. Or are you saying that it does say something about that? If so, I think I missed that in my studies. At the very least, I don't remember it and maybe you could point me to a place where it shows that.

 

[pka]

Oh, but you would agree that "The fact that f(x) is continuous on [a,b] has NOTHING TO DO WITH THE PARTICULAR VALUES OF f(x) on [a,b].

You added the word particular. It was not in the original reply. So you are adding words that are not there.

BTW. Did you really study this?