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There were quite a number of errors in the "correct" answers that were listed. Am I being unnecessarily harsh in expecting that math teachers should know the math that they're teaching? I'm no mathematician and I could certainly be wrong in my explanations. I'd appreciate any feedback pointing out any errors that I've made.

Source link for the full question and my full response:

Question on yahoo answers

1 posted on 10/28/2010 9:58:18 AM PDT by Bob
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To: Bob

That is because most don’t know math. I argue all the time that math skills are not close to what they were 2 generations ago when we put a man n the moon and created computers. Now the teach how to build “T” charts in elementary instead of memorizing the times tables etc... There is no foundation. They teach for the standardized tests.


2 posted on 10/28/2010 10:04:24 AM PDT by Resolute Conservative
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To: Bob
The definition in my book says domain is all the x alues and range is all the y values.
That is the most braindead definition I’ve ever heard. Did anybody really print a book containing such nonsense?
3 posted on 10/28/2010 10:11:12 AM PDT by cartan
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To: Bob

You either “get” math or you don’t. I don’t.

Thankfully, we have worked with Saxon and now my daughters “get it”.

But then again, I taught them times tables and touch math for easy calculations.


4 posted on 10/28/2010 10:12:04 AM PDT by netmilsmom (Happiness is a choice.)
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To: Bob

Wow, that math teacher made some unforgivable mistakes. If the teacher is making the mistaks, the student can’t learn. Like in question 1, for example, the range cannot be the real set {R}, because 0 is excluded ( f(x) =/ 0 for any value of x).

4) is also wrong, because the absolute value of |x-2| will always be a positive number, so the range is from -infinity to 4.


5 posted on 10/28/2010 10:12:07 AM PDT by domenad (In all things, in all ways, at all times, let honor guide me.)
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To: Bob
for the luvofPete, that's not a math question so much as it is a simple definition -- semantics.

For your reading pleasure...Safety in Numbers...

If you're not a one, you're a zero....

6 posted on 10/28/2010 10:12:45 AM PDT by the invisib1e hand (sarc tags are for whimps.)
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To: Bob

It only gets worse. Houston Independent Schools have waivers for students to bring calculators to tests, because the students/parents are too lazy to learn/teach both the addition and multiplication tables.

So we have a generation of High School students who cannot perform simple, basic mathematics without a calculator. The fact that many cannot read, write or speak English is already an accepted fact.


7 posted on 10/28/2010 10:13:00 AM PDT by Hodar (Who needs laws .... when this "feels" so right?)
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To: Bob
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.

The second statement here does not follow from the first. Certainly it is true that both problems should list "f(x) does not equal zero". But that is not because x cannot equal zero.

In problem 2 the sign of f(x) is always positive for x > 0 and negative for x < 0. Since zero is excluded from the domain, f(x) is not zero for any finite values of x. We note that the function gets arbitrarily close to zero as we approach infinity (in both directions) but we assume we are not in the extended real numbers as this is presumably a high school level problem. Problem 1 can be dealt with in a similar manner.

8 posted on 10/28/2010 10:15:26 AM PDT by 17th Miss Regt
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To: Bob

In the first question f(x) = x + 2/x
f(x) does include 0 in this case if x = i * sqrt2
Therefore f(x) does include all real numbers including 0
x cant be 0 because 2/x yields an undefined result

In the 2nd f(x) = 2/x
x cant be 0 as it would yield an undefined result aswell
f(x) cant be zero but it can get approximately close with x being infinite

I miss the days when I was doing math this easy :) Currently doing control systems specifically bode plots and z transforms.. ugh so boring.
And no you arent being harsh on teachers, alot of them are really unqualified in imparting knowledge on their students. They may know the material but they dont know how to teach, I have come across alot of teachers like that both at school and university. Someday when i leave the private sector I want to go into teaching.


9 posted on 10/28/2010 10:16:17 AM PDT by hannibaal
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To: Bob

Actually, the range of y=f(x)=x+2/x is the set of all y whose absolute value is less than or equal to the square root of 8.

You just set 2+2/x=y, turn it into a quadratic equation in x and then compute discriminant, y^2-8.


11 posted on 10/28/2010 10:23:38 AM PDT by Catphish
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To: Bob

It’s all Greek to me

You lost me at f(x)


13 posted on 10/28/2010 10:26:55 AM PDT by libertarian27 (Ingsoc: Department of Life, Department of Liberty, Department of Happiness)
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To: Bob

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.


23 posted on 10/28/2010 11:14:47 AM PDT by PeteCat
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