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I was practicing on some exercises because I have quiz tomorrow, and I got stuck at this exercise, so I wish that someone would help me. here is the exercise.

Given

$$g(x) = \frac{x^2 -16}{x-4}$$

1)Is $g$ continuous at $x=4$? Justify.
2)Does $g$ have a continuous extension at $x=4$? If yes, give this extension.

The first question is easy for me. I have to find that $$\lim_{x\to 4^+} g(x) = \lim_{x\to 4^-} g(x) = g(4)$$

But my problem is the 2nd question because I don't know what continuous extension means and I don't know how to find it either.

Thanks in advance :)

Martin Sleziak
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Ali Zgheib
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1 Answers1

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Suppose $g:D\to Y$ is a function from a set $D\subset X$ to another set $Y$. Then a continuous extension of $g$ is a function $h : X\to Y$ such that $g(x)=h(x)$ for all $x\in D$ and $h$ is continuous. A continuous extension at a point, say $x=4$ means that point is in $X$. I.e. we want the continuous extension to be defined at that point (in this case 4).

In this case how you would find it is just notice that $$\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}.$$ whenever $x\ne 4$, we can cancel and then we have $$\require{cancel} g(x)=\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}}=x+4.$$ So $x+4$ is a continuous extension, since it agrees with the original function wherever $x-4 \ne 0$. i.e. everywhere except $4$ and the original function was not defined there anyway, and $x+4$ is continuous on $\Bbb{R}$.

jgon
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  • Jgon,thanks for your answer. I understand what u mean. i have to find another equation that isn't discontinuous at x=4 .. but i didn't understand how u came out with the continuous extension equation x+4 is there any steps that i have to follow ? And sorry for my bad english :p – Ali Zgheib May 16 '15 at 19:36
  • Oh sorry, let me expand – jgon May 16 '15 at 19:43