If f(0) = 1 and f(2xy - 1) = f(x)f(y) - f(x) - 2y – 1, then f(x) = ?

in Pre-Algebra Answers by Level 1 User (300 points)

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

If f(0) = 1 and f(2xy - 1) = f(x)f(y) - f(x) - 2y – 1, then f(x) = ?

Setting x = 0 in the expression f(2xy - 1) = f(x)f(y) - f(x) - 2y – 1, then,

f(0 - 1) = f(0)f(y) - f(0) - 2y -1

f(-1) = f(y) -1 - 2y - 1

f(-1) = f(y) - 2y - 2

or f(y) = f(-1) + 2y + 2

setting y = 0 in the above equation,

f(0) = f(-1) + 0 + 2

f(-1) = 1 - 2 = -1

f(-1) = -1

Using this value for f(-1), we can rewrite the equation for f(y), to give

f(y) = -1 + 2y + 2

f(y) = 2y + 1

Therefore f(x) = 2x + 1

by Level 11 User (81.5k points)

Related questions

1 answer
asked Mar 22, 2021 in Algebra 1 Answers by anonymous | 725 views
1 answer
asked Aug 12, 2014 in Other Math Topics by Kiroro | 4.0k views
1 answer
1 answer
asked Oct 22, 2019 in Other Math Topics by anonymous | 1.2k views
0 answers
1 answer
1 answer
asked Jan 31, 2014 in Other Math Topics by angel | 1.1k views
1 answer
asked Jun 7, 2013 in Algebra 1 Answers by anonymous | 774 views
1 answer
asked Feb 21, 2013 in Algebra 1 Answers by anonymous | 1.0k views
1 answer
asked Feb 7, 2013 in Algebra 1 Answers by anonymous | 544 views
1 answer
asked Nov 2, 2012 in Algebra 1 Answers by anonymous | 686 views
1 answer
asked Mar 9, 2012 in Algebra 1 Answers by anonymous | 728 views
1 answer
asked Oct 17, 2011 in Algebra 1 Answers by anonymous | 1.9k views
2 answers
asked Sep 26, 2011 in Algebra 2 Answers by anonymous | 1.5k views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,516 questions
100,281 answers
2,420 comments
733,582 users