Finding discontinuities in functions

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1. Feb 28, 2017

Schaus

1. The problem statement, all variables and given/known data
Where are the following functions discontinuous?
f(x) = (x+2)/√((x+2)x)

2. Relevant equations

3. The attempt at a solution
f(x) = (x+2)/√((x+2)x)

= (x+2)/x√(2x) multiply both denominator and numerator by √(2x)

= (x√2+2√x)/(x(2x))
Can I leave it like this and state that x ≠ 0, or have I gone horribly wrong? Otherwise, I'm a little lost. The other questions similar to this that I have done usually consist of factoring out and cancelling a binomial in the numerator with a binomial in the denominator.

Last edited: Feb 28, 2017
2. Feb 28, 2017

Ray Vickson

Do you mean
$$f(x) = x + \frac{2}{\sqrt{x(x+2)}},$$
or do you mean
$$f(x) = \frac{ x+2}{\sqrt{x(x+2)}}?$$
Interpreted literally, what you wrote means the first one.

3. Feb 28, 2017

Schaus

Sorry was meant to be the second one. By the way how do you write questions in that form? Would be much simpler for people to understand what I'm trying to ask if I could display them properly.

4. Feb 28, 2017

Stephen Tashi

You should begin the search for discontinuities in a function that is defined by an algebraic expression by looking at the expression in its original form. Look for values of $x$ that would cause division by zero. Look for values of $x$ that would cause a function like $\sqrt{()}$ or $\log()$ to be undefined.

Then you have to understand the conventions used by your course material. In particular, is a value of $x$ for which the algebraic expression is not a defined supposed to be considered a "discontinuity" of the function or is it merely to be considered a value that is "not in the domain of the function" ?

Notice that you can't do such a manipulation unless $\sqrt{2x}$ is a number. Assuming you are dealing only with the real number system, you would have to assume $2x$ was not a negative number. When trying to do deductive reasoning with algebraic manipulations, you must keep track of such assumptions as you proceed. You can't just wait until to the very end of the process to deduce what restrictions apply to $x$.

The type of manipulations you are doing may be relevant to determining whether a discontinuity is a "removable discontinuity". Have you studied removable discontinuities?

5. Feb 28, 2017

Schaus

So before I do anything I should state the restrictions on the denominator? Which are x ≠ 0 and x ≡ -2, I believe.

6. Feb 28, 2017

Stephen Tashi

Can $x = -1.3$?

7. Feb 28, 2017

Schaus

No it can not....

8. Feb 28, 2017

Schaus

Do I simply state that it has a discontinuity at x = -2 but isn't removable then because otherwise I'm at a loss.

9. Feb 28, 2017

Stephen Tashi

The only safe course of action is to look at examples from your course materials and infer what to do. Can you quote a similar example for us? My guess is that the problem wants you to consider $lim_{x \rightarrow 0} f(x)$ and $lim_{x \rightarrow -2} f(x)$, perhaps by looking at the two-sided limits in each case. If that's what is expected, then your course materials are being careless about certain technicalities - which sometimes happens in our education.

As I indicated, you'll have to straighten out the particular terminology used in your course materials. Some people might say the function is discontinuous at x = -1.3 because it is "not continuous at x = a". Other people might say only that "x = -1.3 is not in the domain of the function" because they regard a definition of a function by an algebraic expression as implicitly excluding values that make the algebraic expression undefined. They might not say "x = -1.3 is a discontinuity of f(x) " because their definition of "discontinuity" requires that the discontinuity occur at a value in the domain of the function.

If your dealing with course materials that expect you to classify each discontinuity as one of 3 types, "removable", "jump" or "essential" then (I think) this implies that a value that is a discontinuity must be in the domain of the function. However, course materials can be careless, Perhaps the the problem should have defined the function as:

$f(x) = \frac{(x+2)}{\sqrt{(x)(x+2)}}$ when $(x)(x+2) > 0$
and $f(x) = 0$ when $x= -2$ or $x = 0$.

That definition would include $x = -2$ and $x = 0$ as values in the domain of $f$.

10. Feb 28, 2017

Ray Vickson

Just use parentheses, like this: f(x) = (x+2)/√(x(x+2)). The point is that a+b/c means $a + \frac{b}{c}$, so if you want $\frac{a+b}{c}$ you need to put the numerator in brackets: (a+b)/c. Similarly, a/b+c means $\frac{a}{b}+c$, so if you mean $\frac{a}{b+c}$ you need to put the denominator in brackets: a/(b+c). Better still: take a few minutes to learn the basics of LaTeX. For in-line LaTeX formulas you still need to use parentheses, but the results look a lot better: $f(x) = (x+2)/\sqrt{x(x+2)}$.

11. Feb 28, 2017

Schaus

The only other question I have to compare to is f(x) = (x2-x)/(x2-1) which I've factored and eliminated discontinuity. (x(x-1))/(x-1)(x+1). I take out the (x-1) so x ≠ 1 and I am left with x/(x+1). The question above is the first I've encountered with a radical and it has thrown me off, considering I am still trying to grasp this whole limit concept. I will definitely look into in-line LaTeX formulas! Thanks for the info