Question:

What is the range of the function....?

by  |  earlier

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f(x) = 1/(x^2 + 1)^(1/2)

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  1. ♣ y’=-x/(x^2+1)^(3/2) =0, hence x = 0; therefore

    max y=1, hence range is 0<y<=1; a bell shape!

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