Question

Let f : N → ℝ be a function defined as f(x) = 4x² + 12x + 15. Show that f : N → S, where S is the range of f, is invertible. Also find the inverse of f.

Answer 1

The given function is f(x) = 4x2 + 12x + 15.
Let us show that f : N → S is a bijection.
f is one-one:
For any \[x_1 , x_2 \in N\] 

For any

\[x_1 , x_2 \in N\] ,

\[\text { Let } f\left( x_1 \right) = f\left( x_2 \right)\]

\[ \Rightarrow 4 x_1^2 + 12 x_1 + 15 = 4 x_2^2 + 12 x_2 + 15\]

\[ \Rightarrow 4\left( x_1^2 - x_2^2 \right) + 12\left( x_1 - x_2 \right) = 0\]

\[ \Rightarrow 4\left( x_1 - x_2 \right)\left( x_1 + x_2 + 3 \right) = 0\]

\[ \Rightarrow x_1 - x_2 = 0 \left[ \because x_1 + x_2 + 3 \neq 0 \text { for any } x_1 , x_2 \in N \right]\]

\[ \Rightarrow x_1 = x_2 \]

So, f : N → S is one-one.
Therefore, f : N → S, where S is the range of f, is onto.
Hence, f : N → S is invertible.
Let

\[f^{- 1}\] denote the inverse of f.

Then

\[\text { fo f }^{- 1} \left( x \right) = x \forall x \in S\]

\[ \Rightarrow f\left[ f^{- 1} \left( x \right) \right] = x \forall x \in S\]

\[ \Rightarrow 4 \left[ f^{- 1} \left( x \right) \right]^2 + 12 f^{- 1} \left( x \right) + 15 = x\]

\[ \Rightarrow 4 \left[ f^{- 1} \left( x \right) \right]^2 + 12 f^{- 1} \left( x \right) + 15 - x = 0\]

\[ \Rightarrow f^{- 1} \left( x \right) = \frac{- 12 \pm \sqrt{144 - 4 \times 4 \times \left( 15 - x \right)}}{8}\]

\[ \Rightarrow f^{- 1} \left( x \right) = \frac{- 12 \pm \sqrt{16x - 96}}{8}\]

\[ \Rightarrow f^{- 1} \left( x \right) = \frac{- 12 \pm 4\sqrt{x - 6}}{8} = \frac{- 3 \pm \sqrt{x - 6}}{2}\]

\[ \Rightarrow f^{- 1} \left( x \right) = \frac{- 3 + \sqrt{x - 6}}{2} \left[ \because f^{- 1} \left( x \right) \in N, \text { so } f^{- 1} \left( x \right) > 0 \right]\]

Thus, the inverse of f is \[\frac{\sqrt{x - 6} - 3}{2}\].