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Consider f: R to [-5,oo) given by f(x)=9...

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

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In `f: R_+ to [ - 5, oo[, f(x) = 9x^(2) + 6x -5`
Let `x, y in R_+ and f(x) = f(y)`
`rArr 9x^(2) + 6x - 5 = 9y^(2) + 6y- 5`
`rArr 3x ( 3x + 2) = 3y ( 3y + 2)`
`rArr " " x =y`
`therefore f ` is one-one.
Let `f(x) = y ` where `y in [-5, oo[`
`rArr 9x^(2) + 6x - 5 = y`
`rArr ( 3x + 1)^(2) = y + 6 rArr 3x + 1 = sqrt(y+ 6)`
`rArr " " x = ( sqrt( y+ 6) -1)/( 3)`
For each ` y in [-5, oo], x = (sqrt(y+ 6)- 1)/(3) in R_+` is such that
`" " f(x) = f((sqrt(y+ 6) -1)/( 3)) = y`
`therefore f ` is onto.
Therefore, `f` is one-one.
`rArr f ` is invertible and `f(x) = y `
`rArr " " f^(-1) (y) = x `
`rArr " " f^(-1) (y) = (sqrt( y+ 6) - 1)/( 3)`
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