Calculus homework studying for the test: Find K such that

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Homework Help Overview

The discussion revolves around a calculus problem involving the continuity of a piecewise function defined as f(x) with a specific condition to find the value of K that ensures continuity at x=0.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the conditions under which the function can be continuous, questioning the implications of the piecewise definition and the behavior of the function as x approaches 0.

Discussion Status

The discussion includes various interpretations of the problem, with some participants suggesting that K could be any real number, while others challenge this view and seek clarification on the continuity condition. Hints regarding limits and continuity have been provided, but no consensus has been reached.

Contextual Notes

Some participants note the need for clarity in the problem statement, and there are references to limits and continuity conditions that are central to the discussion.

huan.conchito
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Calculus homework studying for the test: Find K such that this is continious

[tex]f(x)=\left\{\begin{array}{cc}k&\mbox{ if }x=0\\2+\frac{Sin(x)}{x} &\mbox{ if }x\neq 0\end{array}\right[/tex]

i have no clue how to find such a k that this is continious
 
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well, from your question I see no reason why I can't choose any [itex]k[/itex] that I like. Are you sure that's all there is to it?

Perhaps you are trying to make [itex]f(x)[/itex] continuous?
 
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U didn't say what the problem wa about.I assume it required that the brach function would be continuous on R...Which means point "x=0",too.

What's the condition for an univariable function defined on a domain [itex]\matcal{D}\subseteq \mathbb{R}[/itex] to be continuous in a point [itex]x \in \mathcal{D}[/itex]...?


Daniel.
 
Thats all the question says "find k such that this is continious"
 
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Then the aswer is [itex]k\in \mathbb{R}[/itex],which means an arbitrary #...

Daniel.
 
Thats all the question says "find k such that"
blah blah

perhaps you should elaborate on the "blah blah," verbatim if possible. I find it highly doubtful that that is what it was asking for in a calculus course.
 
huan.conchito said:
f(x) = {k if x=0 if x == 0
{2+(sinx)/x if x =/= 0

i have no clue how to find such a k
Well, IF the objective is to find "k" such that "f(x)" is continuous, what "k" would satisfy that condition??

HINTS #1 & #2:

[tex]1: \ \ \ \ \lim_{x \longrightarrow 0} \, f(x) \ = \ f(0)[/tex]

[tex]2: \ \ \ \ \lim_{x \longrightarrow 0} \, \left (2 \ + \ \frac{\sin(x)}{x} \right ) \ = \ \color{red} \mathbf{?}[/tex]


~~
 
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Well first like this:

[tex]f(x)=\left\{\begin{array}{cc}k,&\mbox{ if }x=0\\2+\frac{Sin(x)}{x}, &\mbox{ if }x\neq 0\end{array}\right[/tex]

Not too hard, check out the syntax in the editor.

Then just do what Xanthym said to assure continuity.
 
Maybe the function needs to be differentiable...:wink:

Daniel.
 
  • #10
the problem is exactly like salty dog wrote it.
and it says
"Find K such that this is continious"
 
  • #11
[tex]2: \ \ \ \ \lim_{x \longrightarrow 0} \, \left (2 \ + \ \frac{\sin(x)}{x} \right ) \ = \ \color{red} \mathbf{2?}[/tex]


xanthym said:
Well, IF the objective is to find "k" such that "f(x)" is continuous, what "k" would satisfy that condition??

HINTS #1 & #2:

[tex]1: \ \ \ \ \lim_{x \longrightarrow 0} \, f(x) \ = \ f(0)[/tex]

[tex]2: \ \ \ \ \lim_{x \longrightarrow 0} \, \left (2 \ + \ \frac{\sin(x)}{x} \right ) \ = \ \color{red} \mathbf{?}[/tex]


~~
 
  • #12
Incorrect.Check that "sin" limit again.

Then u can get conclude what K needs to be...

Daniel.
 
  • #13
the limit = 0, so K must be 0?
 
  • #14
huan.conchito said:
the limit = 0, so K must be 0?

lim {x->0} sin x/x = 1.
 
  • #15
so k must be 3 ?
 
  • #16
Deleting your previous post and then reposting exactly the same thing is a rather cunning way to get an answer :wink:

Anyways, yes, that's fine.
 
  • #17
Correct. If you are unsure why [tex]\ \ \ \ \lim_{x \longrightarrow 0} \, \left (2 \ + \ \frac{\sin(x)}{x} \right ) \ = \ \ \mathbf{1}[/tex] you might have a look at the Taylor expansion of [tex]sin(x)[/tex] .
 
  • #18
Correct. If you are unsure why [tex]\ \ \ \ \lim_{x \longrightarrow 0} \, \left (2 \ + \ \frac{\sin(x)}{x} \right ) \ = \ \ \mathbf{1}[/tex] you might have a look at the Taylor expansion.

You mean [itex]3[/itex] :smile:.
 
  • #19
We have a thread on the "sinc" limit.It has some good posts.

Daniel.
 

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