Consider f : R₊ → [−5, ∞), given by f(x) = 9x² + 6x − 5. Show that f is invertible with f⁻¹(y).

Hence Find
(i) f⁻¹(10)
(ii) y if f⁻¹(y)=43,

where R+ is the set of all non-negative real numbers.

f : R₊ → [−5, ∞) given by f(x) = 9x²+ 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
Let f(x₁)=f(x₂)
⇒9x₁²+6x₁−5=9x₂²+6x₂−5
⇒9x₁²+6x₁=9x₂²+6x₂
⇒9x₁²−9x₂²+6x₁−6x₂=0
⇒9(x₁²−x₂²)+6(x₁−x₂)=0
⇒(x₁−x₂)[9(x₁+x₂)+6]=0
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
9x²+6x−5−y=0
or
9x²+6x−(5+y)=0 where x∈R₊

As x∈R+ i.e., is a positive real number
x cannot be equal to

Since f: R₊ →[-5,∞)
so y ∈ [-5,∞)
i.e y is greater than or equal to -5
i.e. y ≥-5
y+5  ≥0
⇒ Hence the value inside root is positive
Hence √y +6≥0
⇒ x≥0
Hence x is a real number which is greater than or equal to 0.