Question

Let $f:N\to R$ be a function defined as $f\left(x\right)=4x^2+12x+15$ that $f:N\to S$ where $S$ is range of $f_a$ is invariable. Find the inverse $f\left(x\right)$.

Answer 1

Let $\int \left(x\right)=y$

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

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

$x=\frac{-12\pm \sqrt{144-16\left(15-y\right)}}{8}$

$=\frac{-12\pm \sqrt{16\left(9-15+y\right)}}{8}$

$=\frac{-12\pm 4\sqrt{y-6}}{8}$$\int :N\to s$

$soy\in s$$y\in$ range of

$=\frac{-3\pm \sqrt{y-6}}{8}$

As $x\in n$

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

Now $\int \left(x\right)=\int \left(x_2\right)$

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

$\Rightarrow 4\left({x_1}^2-{x_2}^2\right)+12\left(x_1-x_2\right)=0$

$\Rightarrow x_1=x_2$ or $x_1+x_2+3=0$

So $x_1=x_2$ as $x_1+x_2+3\ne 0$

$\therefore f\text{is}\text{one}\text{row,}\text{\&}\text{hence}\text{invertible}\text{and}f\left(x\right)=\frac{-3+\sqrt{y-6}}{2}$