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Math Help - Inc/Decreasing funtions

  1. #1
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    Inc/Decreasing funtions

    Determine the intervals of increase and decrease of the functions

    y = (3x^2 - 2)/x

    y = (1-x^2)/x


    I cant find out how to get the correct answer for these 2. According to the textbook, the answers for the first are (-inf,0) (0,inf) Increase, no decrease.

    And the 2nd is the same but no increase.

    I cant figure out how to find this.
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  2. #2
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    Quote Originally Posted by Sky12 View Post
    Determine the intervals of increase and decrease of the functions

    y = (3x^2 - 2)/x

    y = (1-x^2)/x


    I cant find out how to get the correct answer for these 2. According to the textbook, the answers for the first are (-inf,0) (0,inf) Increase, no decrease.

    And the 2nd is the same but no increase.

    I cant figure out how to find this.
    Simple. If the function is increasing on an interval its first derivative will be positive. Likewise if the function is decreasing on an interval the first derivative will be negative.

    For example: y = \frac{3x^2 - 2}{x}

    y' = \frac{(6x)x - (3x^2 - 2)(1)}{x^2}

    y' = \frac{3x^2 + 2}{x^2}

    The denominator of this is always positive (unless x = 0), so the numerator is the only thing that determines where the first derivative is positive or negative. So let's find where the numerator is 0:
    3x^2 + 2 = 0

    x^2 = -\frac{2}{3}

    And we immediately see that the numerator is NEVER 0 (for any real x, anyway.) So the function is either always increasing or always decreasing. So pick a convenient x value (x can't be 0 remember!), say x = 1000000 will do. It is easy to see that 3 \cdot (1000000)^2 + 2 is positive.

    Thus the first derivative is always positive, so the function is always increasing on its domain. So:
    (-\infty, 0): \text{Increasing}
    (0, \infty): \text{Increasing}

    The second problem is done in a similar manner.

    -Dan
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  3. #3
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    Quote Originally Posted by Sky12 View Post
    Determine the intervals of increase and decrease of the functions

    y = (3x^2 - 2)/x
    Split fractions,
    y=\frac{3x^2}{x}-\frac{2}{x}=3x-\frac{2}{x}.
    And the domain is x\not = 0.
    ---
    Like usual find the derivative and make it null.
    y'=3+\frac{2}{x^2}=0
    Thus,
    \frac{2}{x^2}=-3
    But this has no solutions.
    Thus the function has no criticial points on continous intervals.

    Take any point to the right of zero (because it is discontinous there and the theorem does not apply, so you need to consider both sides). Say x=1
    And check the derivative (which is always positive). Thus, the function is increasing.

    Take any point to the left of zero and the derivative is also positive.
    So the function is increasing when it is strictly less than zero ( x<0 another notation is x\in (-\infty, 0)). And it is increasing when it is strictly larger than zero ( x>0 another notation is x\in (0,+\infty)).

    Here is a picture.
    (The green line is an slant asymptote).
    Attached Thumbnails Attached Thumbnails Inc/Decreasing funtions-picture6.gif  
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  4. #4
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    Quote Originally Posted by Sky12 View Post
    y = (1-x^2)/x
    Split the fraction,
    y=\frac{1}{x}-\frac{x^2}{x}=\frac{1}{x}-x
    Take derivative,
    y'=-\frac{1}{x^2}-1
    Make it zero,
    \frac{1}{x^2}=-1
    But there is no solution.
    And the derivative is always negative.
    In this example the function is always decreasing.
    Similar to the first.
    Attached Thumbnails Attached Thumbnails Inc/Decreasing funtions-picture7.gif  
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