#f(x)=-(x^2+1), or, (x^2+1)=-f(x)#

Let us note that there is no restriction on #x# as to what and what not values it should take from #RR#, clearly, the Domain #D_f# is #RR#

Next, we have, #AA x in RR, x^2>=0, so, (x^2+1)>=1#, i.e., -f(x)>=1,#

& hence, #-1>=f(x).................(1)#.

Recall that the Range of #f#, i.e., #R_f={f(x) : x in D_f}#

#(1)# means that #R_f sub (-oo,-1]...........(2)#

Our claim is #R_f=(-oo,-1]#

To prove this claim; because of #(2)#, we need show that

#(-oo,-1] sub R_f...........(3)#

For this, let #y in (-oo,-1]#, y arbitrary. Then, to show #(3)#, we need show that #y in R_f#, i.e., we need show that #EE# at least one #x in D_f#, such that, #y=f(x)#

To this end, we define #x=sqrt(-1-y)#, & for this #x#, we will show that #y=f(x)#

#y in (-oo,-1]rArry<=-1rArry+1<=0#

#rArr-(y+1)=-1-y>=0rArrsqrt(-1-y)=x# is defined. And, for this #x#, we have, #f(x)=-x^2-1=-(sqrt(-1-y))^2-1=-(-1-y)-1=1+y-1=y#

This proves #(3)# and hence #R_f=(-oo,-1]#

A bit lengthy, but Enjoyable! Enjoy Maths.!