SOLUTION: f(x) = { 2x + 1, x < 0; 2x + 2, x &#8805; 0 f(t^2+1)

Algebra ->  Rational-functions -> SOLUTION: f(x) = { 2x + 1, x < 0; 2x + 2, x &#8805; 0 f(t^2+1)      Log On


   

Question 1090039: f(x) = { 2x + 1, x < 0; 2x + 2, x ≥ 0
f(t^2+1)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Since t%5E2%2B1 is always greater than zero,
f%28t%5E2%2B1%29=2%28t%5E2%2B1%29%2B2
f%28t%5E2%2B1%29=2t%5E2%2B4
f%28t%5E2%2B1%29=2%28t%5E2%2B2%29