(Hi... I'm Dannielle. 8th grade student taking Algebra1) I'm confused... yet again on this preparation packet that's 10% of my grade... Lovely isn't it? Yet another thing my Algebra1 teacher never got the chance to teach. The question says: "A function of g is described : g(x)=(x-3)2 +6... The domain of g is all real numbers greater than 0. The range of g is all real numbers greater than or equal to - A.) 9, B.) 6, C.) 3, D.) -3... " Okay... correct me if I'm wrong, but does g(x) mean basically the same thing as the f(x) but the function is just labeled g instead of f? Yes? No? Sigh.. Okay. All I could think of doing was starting by factoring out the original equation but I just got x2-3x+15. I'm almost positive that I did that wrong and to be completely honest I'm not exactly sure that was what I was supposed to be doing to get the answer. Any suggestions or help would be greatly appreciated...
g(x)=(x-3)^2 +6
- Thread starter Dannielle
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Dannielle, the domain of g is all real numbers, as it is for all polynomial functions. \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \)That is true unless the instructor and/or problem has placed further restrictions on the domain on g(x) = (x - 3)^2 + 6.Dannielle;334076: "A function of g is described : g(x)=(x-3)[SUP said:
I disagree with lookagain (we disagree almost all the time.)
If the problem is as you say it is, then the domain of g(x) is all positive real numbers because that is how the problem defines the domain. Lookagain is correct that if the problem had imposed no restrictions, then the domain would have been all real numbers. However, the problem does not ask about the domain so this disagreement is trivial. The problem as you have given it asks about the range, not the domain. Notice that
\(\displaystyle (x - 3)^2 \ge 0\ for\ all\ real\ x \implies (x - 3)^2 > 0\ for\ all \ real\ x > 0 \implies (x - 3)^2 + 6 > ?? for\ all\ real\ x?\)
With me so far?
So what is the answer?
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I hardly ever disagree with lookagain - HOWEVER, look at the quotation marks. By the definition of g(x) given in the problem, the domain is x>0. g(x) is NOT DEFINED for x < 0.
Doesn't make any difference for the range of g... so ignore us old guys and go ahead and find the minimum of (x-3)^2 + 6, which occurs at x=3.
Okay. I'm maybe... 80% sure that the answer is 3...
D
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No the answer is 6 ← because (x-3)2 has a minimum value of 0.
No, but this one is a little tricky.
When you multiply an inequality by a negative number, the direction of inequality changes.
\(\displaystyle z < 0 \implies (-1) * z > (-1) * 0 \implies (-z) > 0 \implies (-z)^2 > 0^2 = 0 \implies z^2 > 0\ because\ (-z)^2 = z^2.\)
You with me to here?
\(\displaystyle x > 0 \implies 0 < x < 3\ or\ x \ge 3.\)
\(\displaystyle 0 < x < 3 \implies (x - 3) < 3 - 3 \implies (x - 3) < 0 \implies (x - 3)^2 > 0 \implies (x - 3)^2 + 6 > 6.\)
\(\displaystyle x \ge 3 \implies (x - 3) \ge 3 - 3 \implies (x - 3) \ge 0 \implies (x - 3)^2 \ge 0^2 = 0 \implies (x - 3)^2 + 6 \ge 6.\)
\(\displaystyle So\ the\ range\ of\ g(x) = (x - 3)^2 + 6\ is\ all\ real\ numbers\ \ge 6.\)
So... The smallest value of g(x) is when x = 3. Does that mean that 3 is the domain and once you substitute 3 for x and it's all said and done it equals 6. So 6 is the range..? (3-3)(3-3)=0, 0+6=6?
D
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Dannielle,
Do you know what a parabola is (what does one look like)?
Do you have a graphing calculator?
Do you know what a parabola is (what does one look like)?
Do you have a graphing calculator?
No, I don't .
Also, I'm sure that I have a graphing calculator at home but I don't have one with me right now.The concept of a function is so unbelievably simple that kids have a hard time getting it. (At least, I had a hard time getting it back when the dinosaurs made walking to school dangerous.) I suspect that kids say to themselves "A teacher cannot possibly be saying something so obvious and simple so I must be dense."
To be fairly general, a function is an unambiguous rule that says what mathematical operations to perform on a vector. You put a vector in and get a vector out. What is a vector? Well, in the simplest case, it is just a number. So, in the simple case that you are learning, a function is an unambiguous rule that says what arithmetic operations to perform on a number.
f(x) = x + 2 means to add 2 to whatever number is inside the parentheses.
f(x) = 3x means to multiply whatever number is in the parentheses by 3.
f(x) = x means to do nothing to the number in the parentheses.
It's that easy. Why then do we bother with it?
Suppose \(\displaystyle k(x) = \dfrac{(x - 3)^2}{\sqrt{x^2 + 2}} - x^x.\)
It is a whole lot easier to write k(x) than that mess. But it is still just a rule that tells you what operations to perform on x.
Now many times a function is defined for only certain numbers. The numbers for which it is defined are the function's domain. The rule does not apply to numbers outside the function's domain. Sometimes the domain is given explicitly as it was in your problem, where x > 0. Other times, the domain is not given explicitly. If it is not given explicitly, what is meant is that the rule applies to all numbers where the operations required by the rule are permitted. For example, the domain of
\(\displaystyle g(x) = \dfrac{1}{x}\) is all real numbers EXCEPT 0 unless the definition of the function specifies otherwise.
0 is excluded from the domain of the function above because division by 0 is not permitted.
\(\displaystyle h(x) = \dfrac{1}{x}\ for\ x < 0\) has the domain of all negative real numbers because the domain is specified explicitly in the definition of the function.
Got the concept of domain? It is the set of numbers to which the rule applies.
The concept of range is very different. It is the set of numbers which results from applying the function's rule to any number in the domain.
So the range of the function in the problem you gave is any real number greater than or equal to 6. Usually, the range will not be a single number (although it can be).
Any questions?
\(\displaystyle \ \ \ \ \) DrPhil, that problem statement is sloppy/flawed. That is not how it supposed to be presented.Dannielle said:
Either have it as \(\displaystyle \ g(x) \ = \ (x - 3)^2 + 6, \ x > 0, \ \ \ or\)
\(\displaystyle g(x) \ = \ (x - 3)^2 + 6, \ \) with the added restriction that x is greater than 0. \(\displaystyle \ \ \ \ \) Then there would be no domain statement to follow that description of the function.
A problem giver is not to state part of the function, and then in a separate sentence amend what the function is by stating its domain. I did the best I could with what was presented, but I was wrong for doing so, because the problem is stated incorrectly to begin with.
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Sorry
HallsofIvy
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I started algebra, and, to an extent, "functions" when I was in seventh grade. And that was many many years ago.
I would amend the question to "Are functions really taught in 8th grade math?"
I started with 7th grade math in grade 7.
Then I took 9th grade math (elementary algebra) in grade 8.
And then I took 10th grade math (geometry) in grade 9.
And so on.
Today, a person who is 12-years-old is typically beginning grade 7.
And then on up to where the person is 17-years-old is beginning grade 12.
D
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1953 - 12 = 1941
You are old father William....
"You are old, Father William," the young man said,
"And your hair has become very white;
And yet you incessantly stand on your head—
Do you think, at your age, it is right?"
"In my youth," Father William replied to his son,
"I feared it might injure the brain;
But now that I'm perfectly sure I have none,
Why, I do it again and again."
.......
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My mother, celebrating her 75th birthday this Saturday, went to such a school (the elementary school Denis described) in rural Kentucky. She was skipped two grades, going from 2nd grade to 5th. She said the older kids were expected to help the younger kids. Neither of us did algebra until high school. I struggle to keep up with her even now.
My paternal grandfather was the first in his family to finish high school although all three of his younger sisters also finished. He was immediately hired as a teacher in the same high school, in part because he could keep the boys in line. After a few years, he had saved enough to go to college. Not many scholarships back in 1900.
I went to grade school in a three room school. Grades 1-3 in one room. Grades 4-6 in another. And grades 7 and 8 in the last. Looking back, what was great about it was that the kids "tracked" themselves. If you got bored with the 4th grade material, you could always take a shot at the fifth or sixth grade stuff. If you had not quite grasped the fourth grade material when you were in fifth grade, you got remedial lessons with no one the wiser.
I went to grade school in a three room school. Grades 1-3 in one room. Grades 4-6 in another. And grades 7 and 8 in the last. Looking back, what was great about it was that the kids "tracked" themselves. If you got bored with the 4th grade material, you could always take a shot at the fifth or sixth grade stuff. If you had not quite grasped the fourth grade material when you were in fifth grade, you got remedial lessons with no one the wiser.