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Math Help - range of values

  1. #1
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    range of values

    I can't remember how to find the range of values of a function to save my life!

    e.g.
    Given that 2x+3y-5=0, find the range of values of x if y<6.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by colloquial View Post
    I can't remember how to find the range of values of a function to save my life!

    e.g.
    Given that 2x+3y-5=0, find the range of values of x if y<6.

    Thanks in advance!
    One way:
    2x + 3y - 5 = 0

    So
    y = -\frac{2}{3}x + \frac{5}{3}

    So you want to solve
    -\frac{2}{3}x + \frac{5}{3} < 6
    for x.

    -Dan
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  3. #3
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    also the question:
    Find the range of values of p if 3p^2 + 2p >= 5
    so I did:
    3p^2 + 2p - 5 >=0
    (3p+5) (p-1)>=0
    is that even correct?
    :S

    then I would have to draw the number line right?

    thanks again!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by colloquial View Post
    also the question:
    Find the range of values of p if 3p^2 + 2p >= 5
    so I did:
    3p^2 + 2p - 5 >=0
    (3p+5) (p-1)>=0
    is that even correct?
    :S

    then I would have to draw the number line right?

    thanks again!
    Here you need to be a bit more careful.

    3p^2 + 2p \geq 5

    3p^2 + 2p - 5 \geq 0

    (3p + 5)(p - 1) \geq 0

    Now split the number line into 3 pieces: \left ( -\infty, -\frac{5}{3} \right ), \left ( -\frac{5}{3}, 1 \right ), (1, \infty)
    (based on where the LHS is 0) and test your inequality on each of these intervals. The intervals that work give you your solution set.

    In this case:
    \left ( -\infty, -\frac{5}{3} \right ): (3p + 5)(p - 1) \geq 0

    \left ( -\frac{5}{3}, 1 \right ): (3p + 5)(p - 1) \leq 0

    (1, \infty): (3p + 5)(p - 1) \geq 0

    So your solution set will be
    \left ( -\infty, -\frac{5}{3} \right ] \cup [1, \infty).
    (I have included the end points because of the \geq relation. If this had been a > you would not include these points.)

    -Dan
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