Question

Let $f:N\to Y$ be a function defined by $f\left(x\right)=4x^2+12x+15$, where $Y=$ range of $f.$ Show that$f$ is invertible and find the inverse of $f$.

Answer 1

Solution:-

$f:N\to Y$

$f\left(x\right)=4x^2+12x+15$

A function is invertible if the function is one-one and onto.

Let $x_1,x_2\in N$, such that

$f\left(x_1\right)=f\left(x_2\right)$

$4{x_1}^2+12x_1+15=4{x_2}^2+12x_2+15$

$\Rightarrow ({x_1}^2-{x_2}^2)+3(x_1-x_2)=0$

$\Rightarrow (x_1-x_2)(x_1+x_2+3)=0$

$\because x_1,x_2\in N$

$\Rightarrow x_1+x_2+3\ne 0$

$\therefore x_1=x_2$

Thus, $f\left(x\right)$ is one-one.

The function is onto if there exist x in N such that $f\left(x\right)=y$

$\therefore 4x^2+12x+15=y$

$\Rightarrow 4x^2+12x+(15-y)=0$

Here,

$a=4$

$b=12$

$c=15-y$

From quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have

$x=\frac{-12\pm \sqrt{{12}^2-4\times 4\times (15-y)}}{2\times 4}$

$\Rightarrow x=\frac{-3\pm \sqrt{y-6}}{2}$

$\because x\in N$

$\therefore x=\frac{-3+\sqrt{y-6}}{2}$

$f^{-1}\left(x\right)=\frac{-3+\sqrt{x-6}}{2};x\ge 6$