#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.!
