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Math Help - Critical Points- Does this function not have any?

  1. #1
    Junior Member NAPA55's Avatar
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    Critical Points- Does this function not have any?

    Does the function:

    f(x) = 3 + (1/(x+2))

    Have any critical (relative max/min) points?

    My answer is NO, but I'm not 100% certain this is correct.

    The method we were taught is to find the derivative of the function, set it equal to zero, and find the x values. Then we look at intervals to the right and left of this point, and if they are different (pos to neg or neg to pos), then it is a max or min, respectively.

    Finding the derivative of this function and setting it equal to zero, then solving for x, yields a result of 1. Finding the derivative of values left and right of this point gives 2 positive answers, meaning that 1 is NOT a critical point, and therefore there are no critical points.

    The graph has a vertical asymptote at -2 as the function approaches from the left and the right.

    The graph has a horizontal asymptote at 3 as well.

    Both of these conditions seem to imply that there are no maximum or minimum points.

    Yes? No?
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by NAPA55 View Post
    Does the function:

    f(x) = 3 + (1/(x+2))

    Have any critical (relative max/min) points?

    My answer is NO, but I'm not 100% certain this is correct.
    3+\frac{1}{(x+2)^2}

    The derivative is:

    \frac{-2}{(x+2)^3}

    Set it equal to zero:

    \frac{-2}{(x+2)^3} = 0

    You cannot solve for x, hence there being no critical points.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by NAPA55 View Post
    f(x) = 3 + (1/(x+2))
    There is a kind of critical point at x = -2. Critical points are also where the function or first derivative don't exist. A point where the function exists, but the first derivative does not can be either a max or a min.

    -Dan
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  4. #4
    Junior Member NAPA55's Avatar
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    Interesting... thank you both!
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