let f : N - R be a function such that f((x)=4x2+12x+15 Show that f :N - S,where - Maths -

Question

let f : N -> R be a function such that
f((x)=4x2+12x+15 Show that f :N -> S,where S is the
range of f,is invertible Also find f -1

Ajanta Trivedi · Accepted Answer

to show that function is invertible we will show that function is one-one and onto the given function is: $f : N \to R, f (x) = 4 x^{2} + 12 x + 15$ let $x_{1}, x_{2} \in N$ such that $f (x_{1}) = f (x_{2}) 4 {x_{1}}^{2} + 12 x_{1} + 15 = 4 {x_{2}}^{2} + 12 x_{2} + 15 4 {x_{1}}^{2} + 12 x_{1} = 4 {x_{2}}^{2} + 12 x_{2} 4 ({x_{1}}^{2} - {x_{2}}^{2}) + 12 (x_{1} - x_{2}) = 0 ({x_{1}}^{2} - {x_{2}}^{2}) + 3 (x_{1} - x_{2}) = 0 (x_{1} - x_{2}) (x_{1} + x_{2} + 3) = 0$ $x_{1} + x_{2} + 3 \neq 0 \sin c e x_{1}, x_{2} \in N$ thus $x_{1} = x_{2}$ therefore function is one-one the function f is onto if there exist x in N such that f(x)=y $4 x^{2} + 12 x + 15 = y 4 x^{2} + 12 x + 15 - y = 0 x = \frac{- 12 \pm \sqrt{12^{2} - 4 \cdot 4 (15 - y)}}{8} = \frac{- 12 \pm \sqrt{144 - 16 (15 - y)}}{8}$ $x = \frac{- 12 \pm \sqrt{16 y - 96}}{8} = \frac{- 12 \pm 4 \sqrt{y - 6}}{8} x = \frac{- 3 \pm \sqrt{y - 6}}{2} x \in N \Rightarrow x = \frac{- 3 + \sqrt{y - 6}}{2}$ $f^{- 1} (x) = \frac{- 3 + \sqrt{x - 6}}{2} x \geq 6$ hope this helps you

let f : N -> R be a function such that f((x)=4x2+12x+15.Show that f :N -> S,where S is the range of f,is invertible.Also find f -1.

let f : N -> R be a function such that f((x)=4x²+12x+15.Show that f :N -> S,where S is the range of f,is invertible.Also find f ^-1.