Question

se f(x)=x^2+1, encontre e simplifique:
a) f(t+1)
b) (2)
c) [f(t)]^2+1

Answer 1

$\boxed{ f(x)=x^2+1} \\ \\ \boxed{a) f(t+1)=(t+1)^2+1=t^2+2t+1+1=t^2+2t+2} \\ \\ \boxed{b) f(2)=2^2+1=4+1=5} \\ \\ \boxed{c) [f(t)]^2+1=(t^2+1)^2+1=t^4+2t^2+1+1=t^4+2t^2+2}$