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.!
