Question:

Range of the function?

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h(x) = 1/sqrt(2-x) + 5

It's (5,inf), right? Why not [5,inf)? Is there something I should do to get this answer besides the obvious +5 at the end of the equation?

Thanks!

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1 ANSWERS


  1. x cannot be greater than or equal to 2, but it can be very close to 2.

    As x gets close to 2, 1/sqrt(2-x) would get very large (approach infinity).  So f(x) approaches infinity.

    As x becomes more negative (towards negative infinity), 1/sqrt(2-x) would approach zero.  So f(x) would approach 5, but never be exactly 5.  That's why you don't have [5, infinity).

    The range is therefore (5, infinity).

    Hope that helps.

    Take care,

    David

    http://www.tutor-homework.com

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