To prove f is invertible, it is sufficient to prove f is one - one onto.
Here,
f(x) = 5x2 + 6x - 9
One - one :
Let x1,x2 ∈ R,. then,
f(x1) = f(x2)
⇒ 5x12 + 6x1 - 9 = 5x22 + 6x2 - 9
⇒ 5x12 + 6x1 - 5x22 - 6x2 = 0
⇒ 5(x12 - x22) + 6(x1 - x2) = 0
⇒ 5(x1 - x2)(x1 + x2) +6(x1 - x2) = 0
⇒ (x1 - x2) (5x1 + 5x2 + 6) = 0
⇒ x1 - x2 = 0 [ ∵ 5x1 + 5x2 + 6 ≠ 0]
⇒ x1 = x2
i.e., f is one - one function.
Onto :
Let f(x) = y
∴ y = 5x2 + 6x -9
⇒ 5x2 + 6x - (9 + y) = 0
\(⇒x = {-6 \pm \sqrt{36\,+\,4\,\times\,5(9\,+\,y)} \over 10}\)
\(⇒x = {-6 \pm \sqrt{216\,+\,20y} \over 10}\)
\(⇒x = {\pm \sqrt{54\,+\,5y\,-\,3} \over 5}\)
\(⇒x = \frac{\sqrt{54+5y}\,-\,3}{5}\) [ ∵ x ∈ R+]
Obviously,
∀ y ∈ [-9, ∞] the value of x ∈ R+
⇒ f is onto function.
Hence, f is one - one onto function, i.e., invertible.
Also, f is invertible with
\(f^{-1}(y) = \frac{\sqrt{54+5y}\,-\,3}{5}\)