Question

Find the domain and range of the real function, f(x)= 3/1-x square

Answer 1

Answer:

We have, f(x)=1x+3.

Clearly, f(x) is defined for all real values of x except that at which x+3=0,i.e.,x=−3.

∴dom(f)=R−{−3}.

Let y=f(x). Then,

y=1x+3⇒x+3=1y

⇒x=(1y−3). ....(i)

It is clear from (i) that x assumes real values for all real values of by except y =0.

∴range (f)=R−{0}.

Hence, dom (f)=R−{−3}and range(f)=R−{0}.