[Math] Can you solve these problems?

[Klutzington]

Hello, currently I am not impressed with a test grade I received from my math class. I do not find fault in how I got my answer(s), so can you guys show me how you would do these problems? Thanks.

1) f(x) = 11-√(7x-3) ---- Find the domain and range of this function.

2) x^2 [1/2(9-x^2)^-1/2 (-2x)] + (9-x^2)^1/2 (2x) ----- Factor and simplify into one algebraic expression.

3) Given f(x) = 1/(x-2) and g(x) = (x+2)/x. Find and simplify F of G of X OR (f o g)(x)

Thank you very much to each of you who answer. These are the questions I got wrong.


MY ANSWERS:

1) Domain: (-∞, 124/7) (124/7, ∞). Range: (-∞, 0) (0, ∞)

2) -x^3 + 2x (I reviewed this problem and now think the answer is --- (-x^2)/2 ---)

3) 2/(-2+x)

Sorry if this seems like a "do-my-homework" thread. I am going to go back to my math teacher in a few days (he isn't here often), and ask him to clarify.

[Deleted]

Ask the teacher, also, with the lack of your own answers it looks like a do-my-homework thread.

[Klutzington]

I can provide answers. One sec.

Edit: done.

[Verain]

I think you need to list what you answered as well.

Also specify what math class this is. For instance, under imaginary numbers your range and domain has no limit, but with just a real domain your range will never be higher than 11.

For (2), you'll need several lines. Did you try taking everything into pieces, for instance, (9-x^2) becomes (3-x)(3+x)? You can probably put a bunch of stuff under the same power that way, and manipulate them further.

[Klutzington]

I am in pre-calculus. The teacher dubs this "Algebra for Calculus."

[Krekko]

Looks like someone facerolled all over the keyboard to me! Sorry I'm of no help.

[Marthenil]

Just dropping to say this: OH THE MEMORIES D: You just reminded me of high school :<

EDIT: Sorry for not providing any answers, I'm sure I know these, but it's been 3 loong years and I haven't touched math yet on College D:

[Grym]

Something I wanna add about 1)

1) due to √(7x-3), 7x cannot be greater than 3 (unless you worked with Complex number already?), since a negative number times itself will become a positive number eg. -1 x -1 = +1, so unless you include Complex number you cannot have √(-1), so x needs to be equal to or greater than (3/7) as a minimum range.

[Darthcullen]

I got x/(-x+2) for the 3rd one in my head just now. Sorry but i only feel like doing that one

[Hinalover]

For 2, if I am reading this right are you asking:
x^2 ((1/2)((9-x^2)^-1/2)(-2x)) + ((9-x^2)^1/2) (2x)

If so:

x^2((-x)(1/√(9-x^2))) + (2x)(√(9-x^2))
x^2(-x/√(9-x^2)) + 2x(√(9-x^2))
(-x^3/√(9-x^2)) + 2x(√(9-x^2))
2x(√(9-x^2)) - (x^3/√(9-x^2))

If √(9-x^2) is not on the same line then you cannot cancel them out.

A good site for this is Wolframalpha

Link to the problem

[Powerogue]

Gah I recognize the math but am already rusty on it.

#3 I tried doing, I thought "f o g" was "put the f equation in for x in the g equation", I did that and got 2/1.

#1 I'd normally just plug into the graphing calculator and look at the line it forms. There's rules to tell how doing individual things to an equation effects the line, but they're ridiculously complicated, and I forgot them two days after we stopped using them.

#2 Good God... You're on your own. What even happens to a number when you take it to the negative one half power?

[Garnier Fructis]

1) The domain is the possible values of "x" that will satisfy your equation. And, as Grym said, you will get the following answer: The domain is the set of all real numbers x such that x is greater than or equal to 3/7. The range is all possible values of "y." Consider that the smallest value of x (which is 3/7) will cause the square root to be 0, and the answer to the equation to be 11. Every value of x that is greater will cause the answer to be less than 11. Thus, you can conclude that your range is the set of all real numbers less than or equal to 11.

[Klutzington]

#2 Good God... You're on your own. What even happens to a number when you take it to the negative one half power?

Anything to the negative one half (1/2) power means that you put 1 over it and square root it. If it is to the 1/4 power you 4th root it. If it's -1/4 power, it's all under 1 in a fraction bar, to the 4th root.

Believe me, I hate my teacher and I am explained this content much better with my previous math teacher.

[Frah]

for 1
f is valid when
7x-3 > 0
7x > 3
x > 3/7

So f has domain (3/7, infinity)


Since f is linear and decreasing then the range is (-infinity , f(3/7) ) = (-infinity , 11)


For 2 i dunno what the equation is you are wanting. If im thinking its what you wanted to write then im getting
3x(6-x^2) / (9-x^2)^1/2


For 3
f(g(x)) = 1/ (g(x) -2 ) = 1 / ((x+2)/x) - 2x/x) = ( 1 / (2-x) ) (1/x) = 1 / (x(2-x))

[Klutzington]

1) The domain is the possible values of "x" that will satisfy your equation. And, as Grym said, you will get the following answer: The domain is the set of all real numbers x such that x is greater than or equal to 3/7. The range is all possible values of "y." Consider that the smallest value of x (which is 3/7) will cause the square root to be 0, and the answer to the equation to be 11. Every value of x that is greater will cause the answer to be less than 11. Thus, you can conclude that your range is the set of all real numbers less than or equal to 11.

I understand completely. I know how to find "y" (plug and chug "x" after you found it), but how did you guys get the 3/7 part? Simply because it is under the square root? How I ended up with 124/7 was I set the equation equal to 0 (apparently I'm wrong?). Thanks

[Garnier Fructis]

I understand completely. I know how to find "y" (plug and chug "x" after you found it), but how did you guys get the 3/7 part? Simply because it is under the square root? How I ended up with 124/7 was I set the equation equal to 0 (apparently I'm wrong?). Thanks

You don't want to set the whole equation to 0, because only the square root places any limits on acceptable inputs. We know that a square root can't evaluate negative numbers, unless you are working with complex numbers. So what you really want to do is consider this inequality:

7x-3 >= 0
7x >= 3
x >= 3/7

What this tells us is the following: Any number greater than or equal to 3/7 will give you a non-negative value for the square root. Any number below that can't be evaluated. This effectively sets the limit for your domain.

[Hinalover]

Yea for #2, I'm still getting:

2x(√(9-x^2)) - (x^3/√(9-x^2))

You cannot break down √(9-x^2) down further since in order for you to have (9 - x^2); you need (3 + x)(3 - x).

If you REALLY want to break it down even further I would go so far as to say:

x (2(√(9 - x^2)) - ((x^2) / (√(9 - x^2))))

[Klutzington]

You don't want to set the whole equation to 0, because only the square root places any limits on acceptable inputs. We know that a square root can't evaluate negative numbers, unless you are working with complex numbers. So what you really want to do is consider this inequality:

7x-3 >= 0
7x >= 3
x >= 3/7

What this tells us is the following: Any number greater than or equal to 3/7 will give you a non-negative value for the square root. Any number below that can't be evaluated. This effectively sets the limit for your domain.

That makes sense. However I have been working with complex numbers in other places. Though, I doubt he wants irrational (imaginary) solutions.

Zamg the answers to my graph were so wrong v_v.

[Frah]

I put the answers in a previous post but tbh it bugs the hell out of me to write maths in this bad computer code way. I typed up a solution set hopefully with each step clear to understand.

http://www.9thorder.com/pics/solutions.jpg

I can happily answer any questions.

Spotted a mistake already the 1st part should be an equal to or greater than not a greater than :P

[Verain]

1) Domain: (-∞, 124/7) (124/7, ∞). Range: (-∞, 0) (0, ∞)

The domain is all the X values that can be assumed. Domains do not consist of ordered pairs- A domain is just a set of X values. I would claim that the domain is found by trying to find where the square root becomes negative:
7x-3 >= 0
x >= 3/7
So the domain is all X such that X is greater than three sevenths.
TEST: If X is equal to three sevenths, we find 0. If X is greater than three sevenths, we get a positive number. If X is less than three sevenths, the -3 portion is large enough to make the square root hit a negative number.
I'm not exactly sure where you got the 124. I think that's why your solution is incorrect. I'm also puzzled by your listing of two ordered sets.

The range is all values that Y can be. Again, I would not list that as any kind of ordered pair. Because the square root section must be positive, and can be any positive real number, there is no limit as to how small the range can be. But the smallest the square root can be is 0, and 11-0 is 11. I would say your range is:
All Y such that Y is less than or equal to 11.

2) -x^3 + 2x (I reviewed this problem and now think the answer is --- (-x^2)/2 ---)

Well, I'm not going to poke through all of that, but if you think your original answer is wrong, that would make sense!

3) 2/(-2+x)

I would begin by putting (x+2)/x into f(x) as x, which I think is how these things are done.
That would give me 1/ ([x+2/x] -2).
This has an issue where X = 0- it is asymptotic at that point, because we divide by 0.
It also has an issue where X = 2. At that point, the (X+2)/X term is 2, and then subtract 2 and get 0- this is a divide by 0.

With these two nonlinearities in mind, I would try to simplify by multiplying top and bottom by x:
x/([x+2]-2x) = x/(2-x)

And I would add the caveat that this is only accurate for when X is less than 0, or X is greater than 2, or X is between 2 and 0- because it doesn't yield anything useful at 2 or 0. I could be making some mistake in the simplification for sure, but if you didn't exclude values for X, that's probably what got you dinged, even if you algebra was correct and mine is not.