Let y=f(x) then y=2x^2+x+1.
x^2+x/2+1/2=y/2 (dividing through by 2)
x^2+x/2+1/16+7/16=y/2
(x+1/4)^2=y/2-7/16=(8y-7)/16
x+1/4=±√(8y-7)/4 so x=(±√(8y-7)-1)/4.
If x=g(y) then g(y) the inverse function=(±√(8y-7)-1)/4.
So there are really two inverse functions depending on the sign of the square root term. One of these is:
g(y)=(√(8y-7)-1)/4. The other is h(y)=-(√(8y-7)+1)/4.
CHECK
f(0)=1 and g(1)=0; f(1)=4 and f(4)=(√(32-7)-1)/4=(√25-1)/4=4/4=1.
f(0)=1 and h(1)=-1/2, f(-1/2)=1/2-1/2+1=1, so f(0)=f(-1/2)=1.