Manipulating inequalities

If we know that 0 < |x-3| < 2 then...

a) What does this tell us about the possible values of |x-1|?

b) What does this tell us about the possible values of |2x^2 - 8x + 6|?

For a) this is what I did:

If |x-3| < 2 then |x-1| < 4. Therefore any values greater than =3 but less than 5 will be true for |x-1| < 4

b) |2x^2 - 8x + 6| = |2x-2| |x-3| = 2|x-1| |x-3|.

If 2|x-1|<8 (as in above) then |x-1|<4. Put x=2 for 2|x-1|

2|x-3|<4, then |x-3|<2. Therefore, the solution is x=2. Correct?

I am sure for part a I am correct, but not for part b. If I did do wrong, what did I do wrong in part b?

Re: Manipulating inequalities

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**Barthayn** For a) this is what I did:

If |x-3| < 2 then |x-1| < 4.

Yes, with two corrections. By assumption 0 < |x - 3|, so x ≠ 3 and thus |x - 1| ≠ 2. In addition, |x - 3| is *strictly* less than 2, so x ≠ 1 and thus 0 < |x - 1|. Altogether, 0 < |x - 1| < 4 and |x - 1| ≠ 2. I am also wondering how you got your answer since you did not explain it.

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**Barthayn** Therefore any values greater than =3 but less than 5 will be true for |x-1| < 4

But 1 < x < 3 also make 0 < |x - 1| < 4 true. The question does not ask for the set of x's; it asks for possible values of |x - 1|.

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**Barthayn** b) |2x^2 - 8x + 6| = |2x-2| |x-3| = 2|x-1| |x-3|.

If 2|x-1|<8 (as in above) then |x-1|<4.

I don't understand this. Do you start with 2|x - 1| < 8 or |x - 1| < 4? Why do you conclude that |x - 1| < 4 since you established this is a)?

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**Barthayn** Put x=2 for 2|x-1|

Not clear.

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**Barthayn** 2|x-3|<4, then |x-3|<2. Therefore, the solution is x=2.

Also not clear. The solution is an interval of numbers.

In general, you can't find the range of numbers f(x) * g(x) takes just knowing the ranges for f(x) and g(x). For example, f(x) may have a minimum or a maximum where g(x) has a maximum. In this case, however, this is possible because both |x - 1| and |x - 3| get their maximum in the same point x = 5. (Note that x is strictly less than 5.) I would still recommend drawing the graph of |2x^2 - 8x + 6| over the allowed values of x and finding the image from there.

Re: Manipulating inequalities

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**emakarov** Yes, with two corrections. By assumption 0 < |x - 3|, so x ≠ 3 and thus |x - 1| ≠ 2. In addition, |x - 3| is *strictly* less than 2, so x ≠ 1 and thus 0 < |x - 1|. Altogether, 0 < |x - 1| < 4 and |x - 1| ≠ 2. I am also wondering how you got your answer since you did not explain it.

Do you mean By assumption 0 < |x - 3|, so x ≠ 3 and thus |x - 1| ≠ 1? Also, I got this solution by adding one to both sides until I got the function of |x - 1| in the middle. In this case I added to both sides by two.

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**emakarov** But 1 < x < 3 also make 0 < |x - 1| < 4 true. The question does not ask for the set of x's; it asks for possible values of |x - 1|.

So the answer of Therefore, 0 < |x - 1| < 4 is good enough?

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**emakarov** I don't understand this. Do you start with 2|x - 1| < 8 or |x - 1| < 4? Why do you conclude that |x - 1| < 4 since you established this is a)?

My textbook showed me if |x - 1| < 4 then 2|x - 1| < A. To solve for A you would have to go back to the first equation and multiply everything by 2 to solve for in. In this case it would be 8 would it not?

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**emakarov** Not clear.

When I said Put x=2 for 2|x-1| I meant that since the factor form of the function is |2x-2| |x-3| = 2|x-1| |x-3| then you have to "bond" 2|x-1|. To do so you have to solve the x set and then choose an 'x' within that range. x=2 is in that range so you use what what x=2 for 2|x-1| would give you. In this case it will be 2.

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**emakarov** Also not clear. The solution is an interval of numbers.

Would the solution be 1 < x < 5? If not, how would I solve for the solution. I am not sure how one does this. My professor is confusing with mathematics to me. All this stuff is new to me. I feel like my high school did not prepare me with university math. (Crying)

Re: Manipulating inequalities

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I got this solution by adding one to both sides until I got the function of |x - 1| in the middle. In this case I added to both sides by two.

We have x - 3 < 2 iff x - 3 + y < 2 + y for any number y. So, x - 3 < 2 iff x - 1 < 4. However, |x - 3| < 2 means that the distance of x from 3 on the numerical line is less than 2. One can see that this also means -2 < x - 3 < 2. By adding 2 to all parts, this is equivalent to 0 < x - 1 < 4. This *implies* that |x - 1| < 4, but the converse implication is false. Indeed, |x - 1| < 4 is true for x = -2 or for x = 1, but 0 < x - 1 < 4 is false for these values of x. So, going from |x - 3| < 2 to |x - 1| < 4 does not preserve the set of solutions, unlike going from x - 3 < 2 to x - 1 < 4.

Probably the easiest way to think about absolute values is in terms the distance on the numerical line. If the distance of x from 3 is less than 2, then 1 < x < 5. Note also that by assumption, 0 < |x - 3|. Since the absolute value is always nonnegative, the only information this provides is that |x - 3| ≠ 0, i.e., x ≠ 3. So, the solution to 0 < |x - 3| < 2 is 1 < x < 5 and x ≠ 3. On this set, the distance from x to 1 varies from 0 to 4 excluding the ends and 2. Thus, 0 < |x - 3| < 2 implies that 0 < |x - 1| < 4 and |x - 1| ≠ 2.

Concerning |2x^2 - 8x + 6|, draw the graph of 2x^2 - 8x + 6 over (1, 5) and flip the side that is below the x-axis up. It does seem that the problem wants you to represent |2x^2 - 8x + 6| as 2|x-1| |x-3| and find the range of the left-hand side from the ranges of the two factors on the right-hand side. However, as I said, you can only do this sometimes.

Re: Manipulating inequalities

The hint for b) is using the information found above. (I cut three questions out because I got them) Is the solution this: |2x^2 - 8x + 6| < 16? I got this answer from multiplying out the values of |x-1| & |x-3| and then doubling it. IE |x-1| < 4 and |x-3| < 2 then 2|x-1|*|x-3|< 16. Is this logic correct?

Re: Manipulating inequalities

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**Barthayn** Is the solution this: |2x^2 - 8x + 6| < 16? I got this answer from multiplying out the values of |x-1| & |x-3| and then doubling it. IE |x-1| < 4 and |x-3| < 2 then 2|x-1|*|x-3|< 16.

0 < |x - 1| < 4 and 0 < |x - 3| < 2, so 0 < 2|x - 1| * |x - 3| < 16.

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**Barthayn** Is this logic correct?

In this example, yes, but as I said, in general, this gives the superset of the values of the product, not necessarily the exact set. For example, if f(x) = x and g(x) = 2 - x, then on the interval [0, 1], we have 0 <= f(x) <= 1 and 0 <= g(x) <= 2. This would imply that 0 <= f(x)g(x) <= 2, which is true. However, 0 <= x(2 - x) <= 1 on [0, 1].

Re: Manipulating inequalities

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**Barthayn** The hint for b) is using the information found above. (I cut three questions out because I got them) Is the solution this: |2x^2 - 8x + 6| < 16? I got this answer from multiplying out the values of |x-1| & |x-3| and then doubling it. IE |x-1| < 4 and |x-3| < 2 then 2|x-1|*|x-3|< 16. Is this logic correct?

Yes that is correct.

You are saying **if** then the number, **x**, is less than two units from *3*.

Thus if we know that *x* is within two units of *3* then that *x* must be within *4* units of *1*.

In other words .

Re: Manipulating inequalities

Sorry if I sound repetitive, but the solution for the part b would be 0 < |2x^2 - 8x + 6| < 16, correct? I am asking this because, in my opinion, that my professors did not teach the inequality unit well enough to have great understanding so I am not entirely sure if it is correct way to write it.

Re: Manipulating inequalities

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**Barthayn** Sorry if I sound repetitive, but the solution for the part b would be 0 < |2x^2 - 8x + 6| < 16, correct?

Did you read reply #7?

It says if you start with then

Re: Manipulating inequalities

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**Barthayn** Sorry if I sound repetitive, but the solution for the part b would be 0 < |2x^2 - 8x + 6| < 16, correct?

Yes. If we just had increasing functions on (1, 5) whose ranges were (0, 2) and (0, 4), then the range of the product would be (0, 8). Here we have non-monotonic functions and one of them: |x - 3| has a zero in 2. So, their product would involve a 0, unlike (0, 8). However, |x - 1| is viewed over (1, 4) excluding 3 because the the restriction 0 < |x - 3|. So, |x - 1| |x - 3| is never 0 and their range is (0, 8) after all (so the range of 2|x - 1| |x - 3| is (0, 16)).

All I am saying, is the by multiplying the ends of the interval ranges of two functions, one only gets the some superset of the range of the product, which may be a nice approximation to the actual range.

Re: Manipulating inequalities

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**emakarov** Yes. If we just had increasing functions on (1, 5) whose ranges were (0, 2) and (0, 4), then the range of the product would be (0, 8). Here we have non-monotonic functions and one of them: |x - 3| has a zero in 2. So, their product would involve a 0, unlike (0, 8). However, |x - 1| is viewed over (1, 4) excluding 3 because the the restriction 0 < |x - 3|. So, |x - 1| |x - 3| is never 0 and their range is (0, 8) after all (so the range of 2|x - 1| |x - 3| is (0, 16)).

All I am saying, is the by multiplying the ends of the interval ranges of two functions, one only gets the some superset of the range of the product, which may be a nice approximation to the actual range.

Thank you for all you help. This seems like the entire unit seems to be making more sense now.