# Thread: Undefined functions + square roots of negative numbers

1. ## Undefined functions + square roots of negative numbers

Simply put I understand the concept of functions and the concept of square roots of negative numbers/eulers notation. However, I have one problem I cannot make sense of that combines the two. I even have the solution but I do not understand the logic behind it. Any help would be wonderful.

Problem:
If f(x) = √((x^2)-4), then f(x) is undefined for which of the following values of x?

(A) -6
(B) -4
(C) -2
(D) 0
(E) 2

Solution Provided:

0≤x^2
4≤x^2
2≤x or -2≥x

"f(x) is undefined for any values of x that are not within the above range. Since 0 is not within the range, choice (D) is correct."

I must be missing something here. I know most of you are way over qualified to answer this question, but any clarification you can offer would be greatly appreciated.

Thanks

2. Because this is a multiple choice type problem, thebn basically you want to know for which value of x we have $x^2-4$ is a negative number (because taking square root of negative number doesn't make sense)

3. Originally Posted by Chienne
Simply put I understand the concept of functions and the concept of square roots of negative numbers/eulers notation. However, I have one problem I cannot make sense of that combines the two. I even have the solution but I do not understand the logic behind it. Any help would be wonderful.

Problem:
If f(x) = √((x^2)-4), then f(x) is undefined for which of the following values of x?

(A) -6
(B) -4
(C) -2
(D) 0
(E) 2

Solution Provided:

0≤x^2
4≤x^2
2≤x or -2≥x

"f(x) is undefined for any values of x that are not within the above range. Since 0 is not within the range, choice (D) is correct."

I must be missing something here. I know most of you are way over qualified to answer this question, but any clarification you can offer would be greatly appreciated.

Thanks
First, I'll bet they didn't say " $0\le x^2$. I'll bet they said " $0\le x^2- 4$ and you miscopied it.

Second, this has nothing to do with "concept of square roots of negative numbers/eulers notation." In fact it wouldn't make sense if it were not clear that we are dealing only with real numbers.

Because all numbers here are required to be real, the quantity inside the squareroot sign must be non-negative: we must have $x^2- 4\ge 0$. Adding 4 to both sides, [tex]x^2\ge 4[/itex].

If x is a positive number then taking square roots of both sides of that [tex]x\le 2[/itex]. If x is a negative number then, just taking the square root is equivalent to dividing by a negative number which reverses the inequality, so [tex]x\le -2[/itex]. 0 is the one number in the list which does not satisfy either of those inequalities.

Of course, as watchman said, for a "multiple choice" question like this, you can just try to evaluate each: (A ) $f(-6)\sqrt{36-4}= \sqrt{32}$,(B)] $f(-4)= \sqrt{16-4}= \sqrt{12}$,(C) $f(-2)= \sqrt{4-4}= 0$,(D) $f(0)= \sqrt{0-4}= \sqrt{-4}$ NOT a real number!, (D) f(-2)= \sqrt{4-4}= 0.