Finding A Constant To A Piece-Wise Function

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Homework Help Overview

The discussion revolves around a piece-wise function defined as g(x) = (x^2 - a^2)/(x - a) for x ≠ a and g(x) = 8 for x = a. The original poster seeks to determine a specific value for 'a' that ensures the function is continuous across the entire real line.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for the two parts of the piece-wise function to yield the same y-value at x = a, and question how to ensure the limit from both sides approaches this value. There is also a query regarding the limit as x approaches a.

Discussion Status

Some participants have suggested that the limit as x approaches a should equal 8, and there is acknowledgment that finding the appropriate value of 'a' could satisfy the conditions for continuity. The conversation reflects a mix of exploration and clarification without reaching a definitive conclusion.

Contextual Notes

There is mention of a warning regarding the forum's rules, which may indicate constraints on how the discussion is conducted or the nature of the responses expected.

Bashyboy
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The function is g(x) = (x^2 - a^2)/(x - a) if x doesn't equal a; and the second part is g(x) = 8 when x = a. The question asks for me to find a specific value for a so that the function might be continuous on the entire real line.

I know that each part of the piece-wise function needs to equal the same y-value, in order for their graphs to overlap; and, also, that the limit needs to approach the same value from each side of this point of possible discontinuity. Is that right? I am just not certain how to go about this. Could someone possibly prod me towards the correct route of solving this problem? Thank you.
 
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So what must

[tex]\lim_{x\rightarrow a} g(x)[/tex]

be??
 
Would it safe to presume that it would be 8? Also, what is this nonsense about receiving a warning? Is this not the calculus forum?
 
Yes, so you must find a such that

[tex]\lim_{x\rightarrow a} \frac{x^2-a^2}{x-a}=8[/tex]
 
Oh, then that would properly satisfy the conditions of continuity--that g(a) exists, which it does, and the limits from both sides are equivalent to what the function is. Thank you very much. We really need a thank you button like that of the MathHelpForum's.
 

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