Consider `f: R_+->[-5,oo)`given by `f(x)=9x^2+6x-5`. Show that f is invertible with `f^(-1)(y)=(((sqrt(y+6))-1)/3)`

Explanation

We have `f: R arrow[-5, infty),` given by


`

f(x)=9 x^{2}+6 x-5`




(i) Take ` f(x_{1})=f(x_{2})`




`Rightarrow 9 x_{1}^{2}+6 x_{1}-5=9 x_{2}^{2}+6 x_{2}-5 `


`
Rightarrow 9(x_{1}-x_{2})(x_{1}+x_{2})+6(x_{1}-x_{2})=0 `


`
...