BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

functions

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Tue Nov 08, 2011 4:42 am

functions

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.
Top
Source: — Problem Solving |
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 am
Location: Utrecht, The Netherlands
Thanked: 318 times
Followed by:52 members

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))
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Tue Nov 08, 2011 4:42 am

by jzebra10 » Sun Nov 27, 2011 11:43 pm
no, the question says for all numbers y such that y doesn't equal 1.
Top
Master | Next Rank: 500 Posts
Posts: 385
Joined: Fri Sep 23, 2011 9:02 pm
Thanked: 62 times
Followed by:6 members

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.

user123321
Just started my preparation :D
Want to do it right the first time.
Top
Post new topic Post Reply
• Page 1 of 1