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Question (math - inequalities)

  1. Jan 25, 2005 #1
    What do you in these cases:

    A function is g(x) = 2 (x-3) ^2 + 4
    Find the range when 0 <= x <= 6

    steps i took:
    0 <= x <=6
    0-3 <= x-3 <= 6-3
    -3^2 <= (x-3)^2 <= 3^2
    9 <= (x-3)^2 <= 9
    22 <= 2 (x-3)^2 + 4 <=22
    :: what did i do wrong?

    correct answer is:
    { y | 4 <= y <= 22 }

    ---------------------
    case#2

    A certain function is defined as f(x)=x^2 + 3
    if -3 <= x <=3, determine the range

    i got {y | 3 <= y <= 9 }
    correct answer is:
    {y | 3 <= y <= 12}

    Thanks in advance
    Preet
     
  2. jcsd
  3. Jan 25, 2005 #2
    The range for a certain function in a certain domain will be the lowest value of y to the highest value. Let's start with case 2.

    You have the lower limit right, the lowest that function will ever be (in the domain) is 3. Now the upper limit, can the function not be 3^2 + 3 = 12?

    For the first case, you are not just looking at the value of y that corresponds to the upper and lower limits of the domain, you are looking for the LOWEST and the HIGHEST y values in that domain. The upper limit you got right. Now the lower limit, is there any number between 0 and 6 that you can use to make you answer lower than 22?
     
  4. Jan 25, 2005 #3

    Gokul43201

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    a>b does not imply [itex]a^2 > b^2 [/itex], unless both are positive.

    Try some examples :
    2 > -3, but 4 < 9
    -2 > -3, but 4 < 9
     
  5. Jan 25, 2005 #4

    Gokul43201

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    The second problem is pretty straightforward. I don't see how you could have got 9 instead of 12. Why don't you show the steps; like you did with the first problem. This will make it easier to diagnose the error.

    And as for the first problem, the approach you used is not the way to solve it. Ask yourself what values of x in the domain maximize and minimize g(x).

    Clearly g(x) is mimimal, for the smallest value that (x-3)^2 can take, which in turn is minimized by finding the smallest possible value that |x-3| can take. Clearly, this happens when x=3. At this value of x, what value does g(x) take ?
     
  6. Jan 25, 2005 #5
    f(x)=x^2 + 3

    domain:-3 <= x <=3
    the way shown in the book is to turn the domain into the range by performing all of the operations in the function (domain -> function -> range)
    ...
    -3 <= x <= 3
    -3^2 <= x^2 <= 3^2
    9 + 3 <= x^2 <= 9+3
    12 <= x^2 <= 12

    I don't even remember what i'm doing... my exam is coming up and I forgot all of this stuff... my answer doesn't even make sense. I don't know how I got the previous answer.

    Is there some algebraic method of doing this or do you have to think every time which value will give you the min/max values? Even looking at the first one, shouldn't the range be 3 <= y <= 22? 3 is between 6 and 0 and gives the lower y value... but the answer in the book says 4. If you want to apply the method where you apply all of the operations of the function, do you have to find like the greatest and least absolute value or something? TiA.
     
  7. Jan 25, 2005 #6

    Gokul43201

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    Please read my post. I've told you why you can't do this.

    Yes, y= g(x) = 4, when x=3. Please read my posts slowly and carefully. You're panicking, and that just doesn't help you one bit. Calm down and start over.
     
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