# Thread: Calc hmwk- extend the fucntion and remove discontinuity

1. ## Calc hmwk- extend the fucntion and remove discontinuity

I don't understand how to remove the discontinuity on this problem:

(x-4)/( (sq. root of x) - 2), x=4

2. Hello,
Originally Posted by h4hv4hd4si4n
I don't understand how to remove the discontinuity on this problem:

(x-4)/( (sq. root of x) - 2), x=4

Let $\displaystyle f(x)=\frac{x-4}{\sqrt{x}-2}$

There's clearly a discontinuity at x=4.

Now, if you find out that $\displaystyle \lim_{x \to 4^+} f(x)=\lim_{x \to 4^-} f(x)=L$, then you can define the function g :

$\displaystyle g(x)=\left\{\begin{array}{ll} f(x) \quad \forall x \neq 4 \\ L \quad \text{for } x=4 \end{array} \right.$

This removed the discontinuity

3. I'm sorry, I don't think I asked the right question, here goes: is x=4 the extended function?

4. Originally Posted by h4hv4hd4si4n
I'm sorry, I don't think I asked the right question, here goes: is x=4 the extended function?
If I understood well, the extended function is g.
The domain of f was any real number except 4.
The domain of g is any real number.

5. Hello, h4hv4hd4si4n!

Remove the discontinuity: .$\displaystyle f(x) \;=\;\frac{x-4}{\sqrt{x} - 2}$

Since $\displaystyle f(4) \:=\:\frac{4-4}{\sqrt{4}-2} \:=\:\frac{0}{0}$, there is a discontinuity at $\displaystyle x = 4.$
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It is the upper half of a "horizontal" parabola
. . with a hole at (4, 4).

Rationalize: .$\displaystyle \frac{x-4}{\sqrt{x}-2}\cdot\frac{\sqrt{x} + 2}{\sqrt{x}+2} \;=\;\frac{(x-4)(\sqrt{x} + 2)}{x - 4} \;=\;\sqrt{x} + 2$

Hence: .$\displaystyle \lim_{x\to4}f(x) \;=\;\lim_{x\to4}(\sqrt{x}+2) \;=\;4$

Therefore: .$\displaystyle f(x) \;=\;\bigg\{\begin{array}{ccc}\frac{x-4}{\sqrt{x}-2} & & x \neq 4 \\ 4 & & x = 4 \end{array}$

. . and this "fills in the hole" . . . The function is continuous.