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Math Help - Value and limits of a function

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    Value and limits of a function

    This problem appeared on the first test in differential calculus. Since it does not involve any calculus I am hoping this is the right place to present it.

    For this problem, consider f(x) = / x + 2 / / (x + 2)
    If f(x) is expressed without absolute value, f(x) = ?

    EVERYONE including me, thinks it should be = 1 The professor says it is equal to 1 or -1.

    Can someone please explain why? We are also suppose to determine the intervals where f(x) is continuous. Again, I do not understand why they are from negative infinity, - 2 union with (-2, infinity)

    Any help would be much appreciated. Frostking
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    Quote Originally Posted by Frostking View Post
    This problem appeared on the first test in differential calculus. Since it does not involve any calculus I am hoping this is the right place to present it.

    For this problem, consider f(x) = / x + 2 / / (x + 2)
    If f(x) is expressed without absolute value, f(x) = ?

    EVERYONE including me, thinks it should be = 1 The professor says it is equal to 1 or -1.

    Can someone please explain why? We are also suppose to determine the intervals where f(x) is continuous. Again, I do not understand why they are from negative infinity, - 2 union with (-2, infinity)

    Any help would be much appreciated. Frostking
    I assume this is actually

    f(x) = \frac{|x + 2|}{x + 2}.


    The wording of the question is very ambiguous. I assume that to "write this without the absolute value sign" means to write an equivalent expression without the absolute value sign - not just to remove the sign.


    Remember that |x| = x if x < 0 and |x| = -x if x \leq 0.

    So here, |x + 2| can equal either x + 2 if x + 2 \geq 0 or -(x + 2) if x + 2 < 0.


    So Case 1: |x + 2| = x + 2.

    f(x) = \frac{x + 2}{x + 2} = 1.


    Case 2: |x + 2| = -(x + 2)

    f(x) = \frac{-(x + 2)}{x + 2} = -1.


    So your professor is correct.
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