Question

( f=F(x)=3 x^{2}+1 )
( Rightarrow y=3 x^{2}+1 )
( Rightarrow y-1=3 x^{2} )
( Rightarrow x^{2}=frac{y-1}{3} )
( Rightarrow x=sqrt{frac{y-1}{3}}=f^{-1}(x)=sqrt{frac{y-1}{3}} )
Replaeing at ( y=x ) we get in
of ( f(x)} )
ie ( f^{-1}(x)=sqrt{frac{x-1}{3}}, 0 )

# 3. A funciton f:R → R is defined as f(x) = 3x2 +1. Then f-1(x) is : (a) dx-1 (c) f-1 does not exist

Solution