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Math Help - Manipulating inequalities

  1. #1
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    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?
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  2. #2
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    Re: Manipulating inequalities

    Quote Originally Posted by Barthayn View Post
    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.

    Quote Originally Posted by Barthayn View Post
    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|.

    Quote Originally Posted by Barthayn View Post
    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)?

    Quote Originally Posted by Barthayn View Post
    Put x=2 for 2|x-1|
    Not clear.

    Quote Originally Posted by Barthayn View Post
    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.
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  3. #3
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    Re: Manipulating inequalities

    Quote Originally Posted by emakarov View Post
    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.

    Quote Originally Posted by emakarov View Post
    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?

    Quote Originally Posted by emakarov View Post
    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?

    Quote Originally Posted by emakarov View Post
    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.

    Quote Originally Posted by emakarov View Post
    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.
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  4. #4
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    Re: Manipulating inequalities

    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.
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  5. #5
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    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?
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  6. #6
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    Re: Manipulating inequalities

    Quote Originally Posted by Barthayn View Post
    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.
    Quote Originally Posted by Barthayn View Post
    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].
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  7. #7
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    Re: Manipulating inequalities

    Quote Originally Posted by Barthayn View Post
    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 0<|x-3|<2 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 |x-1|<4.
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  8. #8
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    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.
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  9. #9
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    Re: Manipulating inequalities

    Quote Originally Posted by Barthayn View Post
    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 0<|x-3|<2 then 0 < |2x^2 - 8x + 6| < 16
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  10. #10
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    Re: Manipulating inequalities

    Quote Originally Posted by Barthayn View Post
    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.
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  11. #11
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    Re: Manipulating inequalities

    Quote Originally Posted by emakarov View Post
    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.
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