Question

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.

Answer 1

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 x₁, x₂ ∈ R+ and x₁ ≠ x₂

Since x₁ and x₂ are positive,

9(x₁ + x₂) + 6 > 0

∴ x₁ − x₂ = 0⇒ x₁ = x₂

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) = 9x² + 6x −5

= 9x² + 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 = 9x² + 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