The current state of math education (vanity)

Yahoo answers ^ | 10/28/2010 | self

[Bob]

On Yahoo answers today, I came across a disturbing question asked by, presumably, a high school math student. Here is the question:

Please, please explain these answers to me!! I am so frustrated!?

No matter how hard I try, I can not understand domain and range!!! The definition in my book says domain is all the x alues and range is all the y values. That's it. Soo, how the heck do you figuire this out?

1. f(x) = x+2/x
Domain: x does not equal zero
Range: f(x) = any real #

2. f(x) 2/x
Domain: x does not equal 0
Range: f(x) does not equal 0

< snip >

Here's my response:

The domain of a function is the set of values to which the x variable can be set.

Since problems 1 and 2 involve dividing by x, the domain does not include the value zero since dividing by zero is an undefined result.

< snip >

The range of a function is the set of values that the function can result in.

In problems 1 and 2, since x can never be zero, neither can the value of the function be zero. I think that both problems should list f(x) does not equal zero.

< snip >

[added] With all of those errors in the listed ranges and domains, it's no wonder that you're confused. Didn't your teacher know that the errors were there? If not, why not? It looks like your math teacher doesn't know much math.

Sorry for the snips. Since this was originally posted on a Yahoo site, I don't want to violate excerpting requirements.

[Catphish]

I’m certain x is restricted to real numbers here. Unless this was an undergrad taking complex variables and in that case he would probably use z not x. You are talking about something else in terms of the range of the function once you allow x to be complex.

[Bob]

You just multiply through by x, so x + 2/x = y becomes x^2+2=yx. Then you get -x^2+yx-2=0. The discriminant of that is y^2-8. Since you are looking for real values that y can be y^2-8 has to be greater than or equal to 0, i.e., the absolute value of y has to be greater than or equal to the square root of 8.

Thanks for the explanation. I think that the 'y' term in there was what confused me. Duh!

[PeteCat]

for the first the range is all real numbers except between -sqrt8 and sqrt8.
try r - real number candidate
r=x+(1/x)
multiply by x
rx=x^2 + 1
0 = x^2 -rx +1
x=(r+- sqrt(r^2 - 8))/2

using the quadratic formula
http://en.wikipedia.org/wiki/Quadratic_equation

the term under the sqrt has to be positive for real values.

[Grams A]

So we have a generation of High School students who cannot perform simple, basic mathematics without a calculator.”

Made a puchase recently of $1.72. Originally gave teen clerk $2.00 which she entered into her register which registered $.28 coming back to me. Then I found two pennies and gave her that assuming she would just give me $.30 back. She absolutely freaked out because she didn’t know what to do and had to call manager over to correct register. He made a big deal about it and stated, “The register must record the transaction exactly as it occurred” and gave me back $.28. So he - probably early 30’s - apparently didn’t know how to figure it out either. Unfortunately I am old enough to well remember when clerks had to calculate your change in their head and count it back to you. No wonder many stores are going to self check-out and credit cards only.