Question

Prove that the function $f:[0,\infty )\to R$ given by $f\left(x\right)=9x^2+6x-5$ is not invertible. Modify the codomain of the function f to make it invertible, and hence find $f^{-1}.$

Answer 1

First of all, let's check the function f(x) for one-one.
Let $x_1,x_2\in [0,\infty )$ such that $f\left(x_1\right)=f\left(x_2\right).$
i.e.,$9x_1^2+6x_1-5=9x_2^2+6x_2-5\Rightarrow 9(x_1-x_2)(x_1+x_2)+6(x_1-x_2)=0$
$\Rightarrow (x_1-x_2)\left\{9\right(x_1+x_2)+6\}=0\Rightarrow (x_1-x_2)=0$ as $9(x_1+x_2)+6\ne 0$ for $x_1,x_2\in [0,\infty )$
$\therefore x_1=x_2$. So, f(x) is one-one.
Now we'll check the function f(x) for onto.
Let $y\in R$, so for any value of $x\in [0,\infty ),y=f\left(x\right)=9x^2+6x-5$
i.e., $y=(3x{)}^2+2(3x).1+1^2-1^2-5=(3x+1{)}^2-6$
i.e., $(3x+1{)}^2=y+6\Rightarrow x=\frac{\pm \sqrt{y+6}-1}{3}$
$\therefore x=\frac{\sqrt{y+6}-1}{3}$ [As $x=\frac{-\sqrt{y+6}-1}{3}\text{/}\in [0,\infty )$
Now for $y=-6\in R,x=-\frac{1}{3}\text{/}\in [0,\infty ).$
Hence f(x) is not onto so, f(x) is not invertible.
Now note that, we have $x\in [0,\infty )$ i.e., $x\ge 0$ so, $\frac{\sqrt{y+6}-1}{3}\ge 0$
$\Rightarrow \sqrt{y+6}\ge 1\Rightarrow y+6\ge 1\Rightarrow y\ge -5.$
So, if f is redefined as $f:[0,\infty )\to [-5,\infty )$ then $f\left(x\right)=9x^2+6x-5$ becomes an onto function.
Thus, f(x) is one-one and onto both if $f:[0,\infty )\to [-5,\infty )$. Hence f is now invertible.
And, $f^{-1}:[-5,\infty )\to [0,\infty )$ is given by $f^{-1}\left(y\right)=\frac{\sqrt{y+6}-1}{3}.$