Find the exact value of c for which f is continuous on its domain

Click For Summary

Homework Help Overview

The problem involves determining the value of c for which the function f is continuous on its domain, specifically at the point x=1. The function is defined piecewise, with one part applicable for x greater than 1 and the other for x equal to 1.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of continuity and the need to evaluate the limit of the function as x approaches 1. There is uncertainty about how to start this evaluation and whether the numerator is correctly identified. Some suggest using L'Hopital's rule, while others note its prohibition for this problem.

Discussion Status

The discussion is ongoing, with participants exploring the implications of continuity and the correct formulation of the function. There is a focus on evaluating limits without certain methods, and multiple interpretations of the function's numerator are being clarified.

Contextual Notes

Participants mention that L'Hopital's rule is not allowed for this problem, which influences their approach to finding the limit. There is also a need for clarity on the function's definition, particularly the numerator.

Charismaztex
Messages
45
Reaction score
0

Homework Statement



Let the function f: [1,infinity)-->R

<br /> <br /> f(x)=\left\{ \begin{array}{rcl}<br /> \frac{(\sqrt{x}-1+x\sqrt{x-1}}{\sqrt{x^2-1}} &amp; \mbox{,}<br /> &amp; x&gt;1 \\ <br /> c, x=1<br /> \end{array}\right.

Find the EXACT value of c for which f is continuous on its domain.

Homework Equations



N/A

The Attempt at a Solution



I have sketched the graph on the computer and the first half (x greater than 1) part seems to be a single curve going to infinity on the right. I am not quite sure how to find c.

Thanks in advance,
Charismaztex
 
Physics news on Phys.org
Well, can you state what it means for a function to be continuous?
 
Ah, wait, continuous would mean that c would have to be the limit of the function as it tends towards x=1. So how do we evaluate the limit of this function? Can't seem to find a place to start...
 
Just curious: is the numerator

<br /> \sqrt{x} -1 + x\sqrt{x-1}<br />

or is it

<br /> \sqrt{x-1} + x\sqrt{x-1}<br />

It makes a difference in the attack and the answer.
 
Hi, thanks for the reply. The numerator is indeed
<br /> <br /> \sqrt{x} -1 + x\sqrt{x-1}<br /> <br />
 
Try using L'Hopital's rule to evaluate the limit of your function as x->1.
 
Oh yes, that brings up another point I forgot to mention. L'Hopital's rule is banned from this question. The only hint I got was to split the function into its sums...
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
1K
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K