#f(x)=-x^2+2x+1=-x^2+2x-1+2=2-(x-1)^2.#

Since, #x# can take any value from RR, we find that the Domain of #f=D_f=RR.#

Now, we notice that, #AA x in RR, (x-1)^2>=0 rArr -(x-1)^2<=0,# and, adding #2#, #2-(x-1)^2<=2,# i.e., #f(x)<=2,# or, #f(x) in (-oo,2].#

Now, the Range of #f =R_f={f(x): x in D_f=RR},#, it will be clear from above , that #R_f sube (-oo,2].#

Our next goal is to show that #(-oo,2] sube R_f.#, which, altogether, will prove that #R_f= (-oo,2].#

To this end, let #y in (-oo,2], y# arbitrary. We will show that, corresponding to this #y, EE# at least one #x in D_f=RR,# such that, #f(x)=y#, meaning that #y in R_f#.

define #x# by, #x=1+sqrt(2-y)#. As #y in (-oo,2], y<=2, (2-y)>=0#, & hence, #x=1+sqrt(2-y)# **does exist in #RR#**

It remains to be shown that #f(x)=y#, which is done as below:-

#f(x)=2-(x-1)^2=2-{(1+sqrt(2-y))-1}^2=2-(sqrt(2-y))^2=2-(2-y)=y.#

Hence, the Proof that #R_f=(-oo,2].#

I hope, this will be helpful! Enjoy maths.!