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)`

Hence Find

1) `f^(-1)(10)`

2) y if `f^(-1) (y) = 4/3`

where R_{+} is the set of all non-negative real numbers.

#### Solution

f: `R_+ -> [-5, oo]` given by `f(x) = 9x^2 + 6x - 5`

To show: *f *is one-one and onto.

Let us assume that *f* is not one-one.

Therefore there exist two or more numbers for which images are same.

For x_{1}, x_{2} ∈ R+ and x_{1} ≠ x_{2}

Since x_{1} and x_{2} are positive,

9(x_{1} + x_{2}) + 6 > 0

∴ x_{1} − x_{2} = 0⇒ x_{1} = x_{2}

Therefore, it contradicts our assumption.

Hence the function *f* is one-one.

Now, let is prove that *f* is onto.

A function *f* : *X* → *Y* is onto if for every *y* ∈ *Y,* there exist a pre-image in *X*.

f(x) = 9x^{2} + 6x −5

= 9x^{2} + 6x + 1 - 6

=(3x + 1)^{2 }- 6

Now, for all x ∈ R^{+} or [0,∞), f(x) ∈ [−5, ∞)

∴ Range = co-domain.

Hence, f is onto.

Therefore, function f is invertible.

Now, let y = 9x^{2} + 6x − 5

1) `f^(-1) (10) = (sqrt(10+6)-1)/3 = (4-1)/3 = 1`

2) `f^(-1) (y) = 4/3`

`:.(sqrt(y + 6) -1)/3 = 4/3`

`=> sqrt(y + 6)-1 = 4`

`=> sqrt(y+6) = 5`

`=> y + 6 = 25`

=> y = 19