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pratu043
 2 years ago
Best ResponseYou've already chosen the best response.1I'm not sure, but I think its (x + 1)^2 + 3(x  1) + 5

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0f(x) = f[(x 1) + 1]

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0yup that's right @pratu043 :)

shahzadjalbani
 2 years ago
Best ResponseYou've already chosen the best response.0explain @pratu043

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0lol look at what i wrote...f(x1 + 1) = f(x + 0) = f(x) since f(x1) = x^2 + 3x + 5 f(x  1 + 1) = (x+1)^2 + 3(x+1) + 5 got the logic?

shahzadjalbani
 2 years ago
Best ResponseYou've already chosen the best response.01 and 1 will be cancelled then there will f(x) I didn't get.

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0hmmm you get that f(x1 + 1) = f(x) right?

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0now let a = x1 f(a) = a^2 + 3a + 5 agree?

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0now like i said f(x) = f(x1 + 1) = f(a+1) therefore f(a + 1) = (a+ 1)^2 + 3(a+1) + 5

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0oops i made a mistake o.O

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0substituting a made me confused _

shahzadjalbani
 2 years ago
Best ResponseYou've already chosen the best response.0which mistake.......

shahzadjalbani
 2 years ago
Best ResponseYou've already chosen the best response.0I am also confused ...........

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0substituting will actually make it complicated..let me start again do you get that f[(x1) + 1] = f(x)?

shahzadjalbani
 2 years ago
Best ResponseYou've already chosen the best response.0yes now I am clear thank you...........

princeofwetlot
 2 years ago
Best ResponseYou've already chosen the best response.0really? lol..great
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