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pratu043Best ResponseYou've already chosen the best response.1
I'm not sure, but I think its (x + 1)^2 + 3(x  1) + 5
 one year ago

princeofwetlotBest ResponseYou've already chosen the best response.0
f(x) = f[(x 1) + 1]
 one year ago

princeofwetlotBest ResponseYou've already chosen the best response.0
yup that's right @pratu043 :)
 one year ago

shahzadjalbaniBest ResponseYou've already chosen the best response.0
explain @pratu043
 one year ago

princeofwetlotBest ResponseYou've already chosen the best response.0
lol 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?
 one year ago

shahzadjalbaniBest ResponseYou've already chosen the best response.0
1 and 1 will be cancelled then there will f(x) I didn't get.
 one year ago

princeofwetlotBest ResponseYou've already chosen the best response.0
hmmm you get that f(x1 + 1) = f(x) right?
 one year ago

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

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

princeofwetlotBest ResponseYou've already chosen the best response.0
oops i made a mistake o.O
 one year ago

princeofwetlotBest ResponseYou've already chosen the best response.0
substituting a made me confused _
 one year ago

shahzadjalbaniBest ResponseYou've already chosen the best response.0
which mistake.......
 one year ago

shahzadjalbaniBest ResponseYou've already chosen the best response.0
I am also confused ...........
 one year ago

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

shahzadjalbaniBest ResponseYou've already chosen the best response.0
yes now I am clear thank you...........
 one year ago

princeofwetlotBest ResponseYou've already chosen the best response.0
really? lol..great
 one year ago
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