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Studentc14
 2 years ago
functions:
If f(x)=x^2x+1, solve:
1) f(x+1)
2) f(a)
3) 2f(x)2
4) f(2x)
Studentc14
 2 years ago
functions: If f(x)=x^2x+1, solve: 1) f(x+1) 2) f(a) 3) 2f(x)2 4) f(2x)

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Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0CAN I post my answers and someone check it?

jaersyn
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, lol, just substitute whatever is in f(whatever is in here)

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.01) x^2+x+3 2) a^2+a+1 (i didn't know if there was any way to simplify this? 3) 2x^2x 4) 2x^22x+1

kmalone99
 2 years ago
Best ResponseYou've already chosen the best response.1all wrong, it three right 2 f(x)2?

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0can you show me how to do them then?

kmalone99
 2 years ago
Best ResponseYou've already chosen the best response.1\[1) f(x+1)= (x+1)^{2}(x+1)+1\] \[=(x+1)(x+1)x1+1\] \[=x ^{2}+2x+1x=x ^{2}+x+1\]

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, I get #1. I just didn't distribute the negative sign.

kmalone99
 2 years ago
Best ResponseYou've already chosen the best response.12) f(a)=\[f(a)=(a)^{2}(a)+1=a ^{2}+a+1\]

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0i got 2... i dont see what went wrong with 3 and i dont know how to simplify 4

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0wait 3 should be 2x^22x?

kmalone99
 2 years ago
Best ResponseYou've already chosen the best response.1great now on four you just did not square the 2

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0oh! so it would be 4x^22x+1 ?

kmalone99
 2 years ago
Best ResponseYou've already chosen the best response.1yes great job you get A+

Studentc14
 2 years ago
Best ResponseYou've already chosen the best response.0yay! lol I just always make silly simplifying mistakes or I miss a sign or something simple like that. Oh well, thanks very much ^_^
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