HI please solve this absolute value equation with interval notation ..

bobbym · 2017-01-25 13:41:33

You are welcome.

Regarding the question if he wishes to state that f(x) does not exist at x = 0 that is okay with me but ask him why the LHS is 0 instead of being undefined.

thickhead · 2017-01-25 20:58:52

thickhead · 2017-01-26 00:01:28

Hannibal lecter wrote:

Hi, please solve this for me I'm in hurry .....

A rapid method of finding valid intervals( whatever may be the number of factors in numerator and denominator) is count the number of critical points on the right side of an interval. if it is even number( including 0) it is valid interval for the relation >0 type.The reason is simple.All the critical points on the right side of the interval give rise to negative valuefactors.but remember all factors must be of type x- k ,not k-x.

Last edited by thickhead (2017-01-26 00:03:49)

zetafunc · 2017-01-26 00:15:51

Hannibal lecter wrote:

I got this question in the exam and please see this picture it was my solution I know it is wrong ..
http://a.top4top.net/p_3900u2w01.jpg

because I forget to take the intersection between [1,∞) and [-2,∞) which will became (-∞,-2] and (-∞,1] {because there is a number which is "2" didn't make the inequality Satisfied that I found it bewteen -2 and 3, so I flip up these two Intervals then the inetersection will be[-2,1] ) and that is what I do always
but I forget to do that and just put the (3,∞)

the question was from 10 marks and the teacher give me 1! just 1 mark ,,, do I deserve this???? I just want to know that if not, what do I deserve??
please tell me

This solution is not correct. There are lots of errors, and you need to understand why.

First of all, you can't just multiply both sides by (x-3): if x ≥ 3 then x-3 is positive, so multiplying both sides by (x-3) won't change the sign. However, if x < 3, then (x-3) is negative, so multiplying both sides by (x-3) will change the direction of the inequality.

Second of all, if you have (x-1)(x+2) ≥ 0, that does not imply that (x-1) ≥ 0 and (x+2) ≥ 0. For instance, we could have (x-1) < 0 and (x+2) < 0. What you have is a quadratic polynomial, with roots at x = 1 and x = -2. Draw a graph of this polynomial and label it to find out when that polynomial goes above the x-axis.

I'm not sure what you are trying to do with the line "(x-3) > 0". If you consider the cases x ≥ 3 and x < 3 separately, you won't need to do this.

thickhead · 2017-01-26 02:55:08

Hi zetafunc,
As i have already posted, there are 3critical points x=-2,x=1 and x=3
the region x>3 has no critical point(even number)on its right. So it is valid region.
again -2<=x<=1 has 2 critical points (even number) on its right. So it is also a valid region.
i think that settles the matter.

bobbym · 2017-01-26 04:33:38

Hi;

Technically, that does not settle the matter. I provided that answer in post #4. The real question the OP is asking appears later.

zetafunc is pointing out some tactical errors that the OP made that I did not because he was very distraught over some of the actions of his teacher.

Now as to the real question, it starts at post #11.

thickhead · 2017-01-26 22:02:25

Hi Bobbym and zetafunc,
I can tell you what Hannibal did. To look methodical he first took the numerator. It was not his intention to multiply the whole by x-3.Then for each term he found the region where it was positive.Then he dealt with the denominator. thereafter he took intersection of all the regions where the terms were found positive.

bobbym · 2017-01-27 02:32:17

Hi;

I did not pay too much attention to his method because he did make a mistake in his answer and we were discussing how much partial credit he should be given. For the actual method, you and zetafunc would be better to explain it to him.

Hannibal lecter · 2017-01-27 10:37:39

bobbym , the teacher said to me when I asked him : we chose the R.H.S = R.H.s and it will be k value = {-1,4}
when only we see have (≥) and (<)
not = and ≠ , cause in (≠ and = )
we chose the equation that have (=) and solve it

he said that to me

Last edited by Hannibal lecter (2017-01-27 10:42:25)

bobbym · 2017-01-27 12:15:39

I am sorry, I do not follow his explanation.

If the limit of the LHS does not exist then why does he say it is 0?

thickhead · 2017-01-27 19:20:12

k={-1,4} does not come unless L.H.S. is taken as 1.The only thing is Lim x->0 k^2-3k-3 is incorrect as k^2-3k-3 is the value of f(x) at x=0. It is not the limit.The concepts come into play while evaluating ,not the answers.

Hannibal lecter · 2017-01-27 21:43:25

I don't know bobbym he even don't know what are he doing, like me and the remain of students....

thickhead I don't know what do you mean be (Lim x->0 k^2-3k-3 is incorrect..)

do you mean my solution is wrong?

Hannibal lecter · 2017-01-27 21:46:33

zetafunc wrote:

Draw a graph of this polynomial and label it to find out when that polynomial goes above the x-axis.
I'm not sure what you are trying to do with the line "(x-3) > 0". If you consider the cases x ≥ 3 and x < 3 separately, you won't need to do this.

Hi, can you please teach me ho to draw a qraph for this? and how to find a full solution in an
actual way?

please we have a two more exams and we always get like these questions ...

thickhead · 2017-01-27 22:41:22

What I say is k^2-3k-3 does not go well with lim x->0. It is the value of f(x) at x=0. Statements do reflect the understanding.I do not think your teacher points at L.H.S. but at R.h.S but if he says k^2-3x-3=0 then he is wrong.

Last edited by thickhead (2017-01-27 22:53:53)

bobbym · 2017-01-28 19:29:06

Based on how he posed the problem that x does not equal 0. The limit of the LHS does not exist at x = 0.

But when he says k^2-3x-3=0, then I can not agree.

thickhead · 2017-01-28 21:45:20

f(x) is defined by f(x)=x^3+1 for x ≠ 0 , and at x=0 f(x)=k^2-3k-3.
So f(0) is not to be calculated as 0^3+1 but to be arbitrarily taken as k^2-3k-3.In general it leaves a removable discontinuity at x=0.But for a limited values of k it could become continuous.this is how I interpret the problem.

Hannibal lecter · 2017-01-29 03:42:33

I talked and discussed to him today but he changed his answer to this! :- see this is my sol in red color and his solution in blue color

He said you must take f(0) = R.H.S then find k not L.H.S=R.H.S,

what is that crazy man talking about???????? even he is right but it is the same result????
and he didn't teach us to do that I swear in god!!!

bobbym · 2017-01-29 07:26:59

Now he is evaluating the answer as I did earlier. You can tell him that is wrong too, or maybe just leave it alone.

Hannibal lecter · 2017-01-29 08:39:26

Please tell me answers that I can tell these answers to him so I can defend my right and myself please explain to me why wrong...
please I need these 10 marks , I need the prove of his wrong

and thank you cause you told me he is wrong now I'm more quietly and relaxing

Last edited by Hannibal lecter (2017-01-29 09:09:55)

bobbym · 2017-01-29 09:53:08

To which part? He is agreeing with your answer now. But remember, that was not your original answer.

Hannibal lecter · 2017-01-29 10:09:16

yes it is .. sorry I don't care for the R.H.S and L.H.S I changed between them every time cause I forget always what is the L.H.S and R.H.S

anyway he said f(0) =L.H.S (it is the K^2+3K-3)
and so it is 1=it is the K^2+3K-3.............,

he said to me it is not Limit R.H.S ( x+1) = Limit L.H.S (K^2+3K-3)

and you said he is wrong too

so how he is wrong in that part?
and it is the same result so why he is doing that

Last edited by Hannibal lecter (2017-01-29 10:11:21)

bobbym · 2017-01-29 10:28:53

Does he say lim x^3 + 1 as x approaches 0 is 1? That is what I said in the beginning. I suggest you agree with him on that.

Hannibal lecter · 2017-01-29 10:42:14

no no no ,

He is said my solution is wrong cause I toke L.H.S = R.H.S to find k value
he said you must write it like this lim L.H.S {which x+1} = f(0) {which it K^2+3k-3)

not Lim L.H.S {which x+1} = Limit R.H.S which it K^2+3k-3}

you do understand what he is saying ...

Last edited by Hannibal lecter (2017-01-29 10:42:51)

bobbym · 2017-01-29 10:55:32

He asked for the limit which is

That is the same as f(0).

Hannibal lecter · 2017-01-29 20:36:29

I told him it is the same, He told you did a crime in mathematics science when you wrote it like that!!!
He want excuses just that..
anyway lets close this section I'll will deal with that problem and I will refer my rights in my way
_____________

I have another question please ...
lets suppose there is x in the question (xk^2+3k-3)
In this position, do I have to take the limit R.H.S or the f(0)?

Math Is Fun Forum

#26 2017-01-25 13:41:33

Re: HI please solve this absolute value equation with interval notation ..

#27 2017-01-25 20:58:52

Re: HI please solve this absolute value equation with interval notation ..

#28 2017-01-26 00:01:28

Re: HI please solve this absolute value equation with interval notation ..

#29 2017-01-26 00:15:51

Re: HI please solve this absolute value equation with interval notation ..

#30 2017-01-26 02:55:08

Re: HI please solve this absolute value equation with interval notation ..

#31 2017-01-26 04:33:38

Re: HI please solve this absolute value equation with interval notation ..

#32 2017-01-26 22:02:25

Re: HI please solve this absolute value equation with interval notation ..

#33 2017-01-27 02:32:17

Re: HI please solve this absolute value equation with interval notation ..

#34 2017-01-27 10:37:39

Re: HI please solve this absolute value equation with interval notation ..

#35 2017-01-27 12:15:39

Re: HI please solve this absolute value equation with interval notation ..

#36 2017-01-27 19:20:12

Re: HI please solve this absolute value equation with interval notation ..

#37 2017-01-27 21:43:25

Re: HI please solve this absolute value equation with interval notation ..

#38 2017-01-27 21:46:33

Re: HI please solve this absolute value equation with interval notation ..

#39 2017-01-27 22:41:22

Re: HI please solve this absolute value equation with interval notation ..

#40 2017-01-28 19:29:06

Re: HI please solve this absolute value equation with interval notation ..

#41 2017-01-28 21:45:20

Re: HI please solve this absolute value equation with interval notation ..

#42 2017-01-29 03:42:33

Re: HI please solve this absolute value equation with interval notation ..

#43 2017-01-29 07:26:59

Re: HI please solve this absolute value equation with interval notation ..

#44 2017-01-29 08:39:26

Re: HI please solve this absolute value equation with interval notation ..

#45 2017-01-29 09:53:08

Re: HI please solve this absolute value equation with interval notation ..

#46 2017-01-29 10:09:16

Re: HI please solve this absolute value equation with interval notation ..

#47 2017-01-29 10:28:53

Re: HI please solve this absolute value equation with interval notation ..

#48 2017-01-29 10:42:14

Re: HI please solve this absolute value equation with interval notation ..

#49 2017-01-29 10:55:32

Re: HI please solve this absolute value equation with interval notation ..

#50 2017-01-29 20:36:29

Re: HI please solve this absolute value equation with interval notation ..

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