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by jzebra10 » Sun Nov 27, 2011 6:16 am
Please help! I always have problems with functions. How would you go about solving this?

for all numbers y such that y doesn't equal 1, if f(y)= y^2-2/y+1. then (1/f(y))(1/f(2)) = ?
Last edited by jzebra10 on Sun Nov 27, 2011 11:41 pm, edited 2 times in total.
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by neelgandham » Sun Nov 27, 2011 10:33 am
for all numbers y such that y doesn't equal 1, if f(x)= y^2-2/y+1. then (1/f(x))(1/f(2)) = ?
Did you mean, For all numbers of y such that y doesn't equal -1, f(y) = (y^2-2)/(y+1), then (1/f(x))(1/f(2)) = ?

f(x) = (x^2-2)/(x+1) (Replace y with x)
f(2) = (2^2-2)/(2+1) = 2/3 (Replace y with 2)

(1/f(x))(1/f(2)) = 1/(f(x)*f(2)) = (3/2)*((x+1)/(x^2-2))
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by jzebra10 » Sun Nov 27, 2011 11:43 pm
no, the question says for all numbers y such that y doesn't equal 1.
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by user123321 » Mon Nov 28, 2011 5:07 pm
jzebra10 wrote:no, the question says for all numbers y such that y doesn't equal 1.
As said by others, y cannot be -1, the function value will become infinite. Seems problem in the problem statement.

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