Q

# Consider f : R + → [– 5, ∞) given by f (x) = 9x ^ 2 + 6x – 5. Show that f is invertible with f -1 (y) = square root of y + 6 - 1 over 3

Q .9 Consider $f : R_+ \rightarrow [- 5, \infty)$ given by $f (x) = 9x^2 + 6x - 5$. Show that $f$ is invertible with $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

Views

$f : R_+ \rightarrow [- 5, \infty)$

$f (x) = 9x^2 + 6x - 5$

One- one:

Let       $f(x)=f(y)$    $for \, \, x,y\in R$

$9x^{2}+6x-5=9y^{2}+6y-5$

$9x^{2}+6x=9y^{2}+6y$

$\Rightarrow$              $9(x^{2}-y^{2})=6(y-x)$

$9(x+y)(x-y)+6(x-y)= 0$

$(x-y)(9(x+y)+6)=0$

Since, x and y are positive.

$(9(x+y)+6)> 0$

$\therefore x=y$

$\therefore$   f is one-one.

Onto:

Let for  $y \in [-5,\infty)$  , $y=9x^{2}+6x-5$

$\Rightarrow$         $y=(3x+1)^{2}-1-5$

$\Rightarrow$             $y=(3x+1)^{2}-6$

$\Rightarrow$                  $y+6=(3x+1)^{2}$

$(3x+1)=\sqrt{y+6}$

$x = \frac{\sqrt{y+6}-1}{3}$

$\therefore$ f is onto  and range is $y \in [-5,\infty)$.

Since f is one-one and onto so it is invertible.

Let  $g : [-5,\infty)\rightarrow R_+$    by  $g(y) = \frac{\sqrt{y+6}-1}{3}$

$(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{{(3x+1)^{2}}-6+6} -1$

$(gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x$

$(fog)(x)=f(g(x))=f(\frac{\sqrt{y+6}-1}{3})$

$=[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6$

$=(\sqrt{y+6})^{2}-6$

$=y+6-6$

$=y$

$\therefore gof=fog=I_R$

Hence,  $f$ is invertible with the inverse $f^{-1}$ of $f$ given by  $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

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