A quick question (1 Viewer)

[Ragerunner]

Solve for x:

Absolute value of (x-1)/(x+1) < 1

Answer is 0 < x < 2

How do I get it?

thanks.

edit: also:

f(x) = squareroot(1 - 2sinx)

Find the maximum domain and range.

 

[speersy]

for the domain
1-2sinx has to be >0
1>2sinx
sinx<1/2 (it is actually less than or equal to)
x<pi/6, x>5pi/6
5pi/6<x<pi/6


For the range
0<f(x)<1

not sure but maybe this is right .

 

[~*HSC 4 life*~]

for question 1 you have 2 cases:
Case 1
x-1/x+1 < 1
or case 2.

x-1/x+1 <-1
then you solve each one as a normal inequality with unknowns in the denominator...

case 1.

x-1/x+1 <1

mult both sides by (x+1)^2

(x-1)(x+1)< (x+1)
(x-1)(x+1)-(x+1)<o
(x+1) [(x-1)-1]<0
(x+1)(x-2)<0

sketch this on a parabola, conc up, x int -1 and 2, we see from graph -1 < x < 2

case 2.
x-1/x+1<-1
(x-1)(x+1)<-(x+1)
(x-1)(x+1)+(x+1)<0
(x+1)[(x-1)+1]<0
(x+1)(x)<0
sketch a parabola, conc up, x int 0 and -1
-1 < x < 0
however x cannot equal to -1 (or else denom is zero and undefined)
therefore 0 < x < 2

 

[Heinz]

Couldnt do the first one for some reason but the second one i got the same answer as speersy. Just make the square root greater than zero and solve for x. And y will always be greater than zero (the square root of any positive number >0 will also be positive)

 

[speersy]

my msg b4 seems to be stuffing up
i will try it again
for question 2

for the domain
1-2sinx has to be >0
1>2sinx
sinx<1/2 (it is actually less than or equal to)

5pi/6 < x < pi/6 (should be less than or equal to signs)


For the range
0 < f(x) < 1

not sure but maybe this is right .

 

[~*HSC 4 life*~]

err i dont know why my final line isnt comign up
therefore
0 < x < 2

 

[Heinz]

Originally posted by ~*HSC 4 life*~
err i dont know why my final line isnt comign up
therefore
0<x<2

Theres a typo in your working "sketch this on a parabola, conc up, x int -1 and -2..." -1 and 2?

 

[speersy]

you have to put gaps in between numbers and > or <
well thats what i had to do anyway

 

[~*HSC 4 life*~]

Originally posted by speersy
you have to put gaps in between numbers and > or <
well thats what i had to do anyway

lol thanks edited

Originally posted by Heinz
Theres a typo in your working "sketch this on a parabola, conc up, x int -1 and -2..." -1 and 2?

thats what i meant- thanks!

 

[Ragerunner]

Thanks for the replies.

I think for the first question the 2 isn't needed, i.e. it was a mistake, i heard someone saying that but i didn't know which question he was talking about.

for the second question the answer shows as:

{x : 2npi < x < (2n + 1)pi }

< is actually less than or equal to.

I don't understand what the hell that "n" is in that answer.

 

[Heinz]

Originally posted by Ragerunner
Thanks for the replies.

I think for the first question the 2 isn't needed, i.e. it was a mistake, i heard someone saying that but i didn't know which question he was talking about.

for the second question the answer shows as:

{x : 2npi < x < (2n + 1)pi }

< is actually less than or equal to.

I don't understand what the hell that "n" is in that answer.

About the first question... i think the answers just x>0 as x-1/x+1 will always be <1 for all x >0. The second question simply means that the domain is inifinite for all n (any integer). Since sinx has to be positive it means that x is between 2npi < x < (2n +1)pi. e.g. for n = 0, its between 0 and pi. Something like that.

Edit: bloody typos

 

[CM_Tutor]

Heinz is correct, the answer to |(x - 1) / (x + 1)| < 1 is x > 0, but there are easier ways to do this than those mentioned above. Remember that a question can penalise you time in an exam by you doing it the long way.

Method 1 - quick algebra.

Since x <> -1 (othewise the inequality is undefined), and |x + 1| is positive otherwise, multiply through by |x + 1|.

The question becomes: |x - 1| < |x + 1|

Taking cases, we have x - 1 < x + 1, which is true for all x, and
-(x - 1) < x + 1
-x + 1 < x + 1
-x < x
0 < 2x
x > 0

Combining these results, we get x > 0.

Method 2 - Quick curve sketching

After getting |x - 1| < |x + 1|, as above sketch the two curves y = |x - 1| and y = |x + 1| on the same set of axes. Both are standard "V" shaped absolute value curves, with x - intercepts at -1 (for |x + 1|) and at 1 (for |x - 1|). These meet at (0, 1).

From these graphs, it shoud be obvious that |x - 1| < |x + 1| when x > 0.

Method 3 - Slower curve sketching.

(x - 1) / (x + 1) = (x + 1 - 2) / (x + 1) = 1 - 2 / (x+1)

Thus, to solve the absolute value inequality, all we need to do is sketch y = 1 - 2 / (x + 1), and look when -1 < y < 1.

This curve is a rectangular hyperbola, with aymptotes of y = 1 and x = -1, and intercepts at (0, -1) and (1, 0).

From the sketch, it should be obvious that the solution is x > 0.

 

[Xayma]

Originally posted by Heinz
Since sinx has to be positive it means that x is between 2npi < x < (2n +1)pi. e.g. for n = 0, its between 0 and pi. Something like that.

Looking at how its setup 2n for any integer just adds a multiple of 2pi to each answer ie it is the general answer since 360sinx=0sinx=720sinx etc

 

[Heinz]

Originally posted by Xayma
Looking at how its setup 2n for any integer just adds a multiple of 2pi to each answer ie it is the general answer since 360sinx=0sinx=720sinx etc

Thats what i meant (im not as articulate as some)

 

[Xayma]

Originally posted by Heinz
Thats what i meant (im not as articulate as some)

Thats what I thought you meant, but it didn't sound right.

 

[~*HSC 4 life*~]

Originally posted by CM_Tutor
Heinz is correct, the answer to |(x - 1) / (x + 1)| < 1 is x > 0, .

would my answer be marked wrong?, i dont get why you disregarded the 2

 

[CM_Tutor]

Originally posted by ~*HSC 4 life*~
for question 1 you have 2 cases:
Case 1
x-1/x+1 < 1
or case 2.

x-1/x+1 <-1
then you solve each one as a normal inequality with unknowns in the denominator...

case 1.

x-1/x+1 <1

mult both sides by (x+1)^2

(x-1)(x+1)< (x+1)
(x-1)(x+1)-(x+1)<o
(x+1) [(x-1)-1]<0
(x+1)(x-2)<0

sketch this on a parabola, conc up, x int -1 and 2, we see from graph -1 < x < 2

The algebra in this is wrong in the first line. Multiplying by (x + 1)^2 actually gives:

(x - 1)(x + 1) < (x + 1)^2
(x - 1)(x + 1) - (x + 1)^2 < 0
(x + 1)[(x - 1) - (x + 1)] < 0
(x + 1)(x - 1 - x - 1) < 0
-2(x + 1) < 0
x + 1 > 0
x > -1

So, there should be no '2' in your case one answer, and you could easily check this by trying a value greater than 2 in the original equation.

Originally posted by ~*HSC 4 life*~
case 2.
x-1/x+1<-1
(x-1)(x+1)<-(x+1)
(x-1)(x+1)+(x+1)<0
(x+1)[(x-1)+1]<0
(x+1)(x)<0
sketch a parabola, conc up, x int 0 and -1
-1 < x < 0
however x cannot equal to -1 (or else denom is zero and undefined)
therefore 0 < x < 2

For the case 2, you really want to solve (x - 1) / (x + 1) > -1, which leads to the answer x > 0 or x < -1.

Then, since the answer must satisfy both cases simultaneously, we see that the answer is x > 0.

So, to answer your question, yes your answer would lose some (but, of course, not all) marks.