# Consider the function f : R+ →[−9,∞] given by f(x) = 5x2 + 6x − 9.

Question:

Consider the function $f: \mathrm{R}^{+} \rightarrow[-9, \infty]$ given by $f(x)=5 x^{2}+6 x-9$. Prove that $f$ is invertible with $f^{-1}(y)=\frac{\sqrt{54+5 y-3}}{5}$ [CBSE 2015]

Solution:

We have,

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

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

$=5\left(x^{2}+\frac{6}{5} x-\frac{9}{5}\right)$

$=5\left(x^{2}+2 \times x \times \frac{3}{5}+\frac{9}{25}-\frac{9}{25}-\frac{9}{5}\right)$

$=5\left(\left(x+\frac{3}{5}\right)^{2}-\frac{9}{25}-\frac{9}{5}\right)$

$=5\left(x+\frac{3}{5}\right)^{2}-\frac{9}{5}-9$

$=5\left(x+\frac{3}{5}\right)^{2}-\frac{54}{5}$

$\Rightarrow y+\frac{54}{5}=5\left(x+\frac{3}{5}\right)^{2}$

$\Rightarrow \frac{5 y+54}{25}=\left(x+\frac{3}{5}\right)^{2}$

$\Rightarrow \sqrt{\frac{5 y+54}{25}}=x+\frac{3}{5}$

$\Rightarrow x=\frac{\sqrt{5 y+54}}{5}-\frac{3}{5}$

$\Rightarrow x=\frac{\sqrt{5 y+54}-3}{5}$

Let $g(y)=\frac{\sqrt{5 y+54}-3}{5}$

Now,

$f o g(y)=f(g(y))$

$=f\left(\frac{\sqrt{5 y+54}-3}{5}\right)$

$=5\left(\frac{\sqrt{5 y+54}-3}{5}\right)^{2}+6\left(\frac{\sqrt{5 y+54}-3}{5}\right)-9$

$=5\left(\frac{5 y+54+9-6 \sqrt{5 y+54}}{25}\right)+\left(\frac{6 \sqrt{5 y+54}-18}{5}\right)-9$

$=\frac{5 y+63-6 \sqrt{5 y+54}}{5}+\frac{6 \sqrt{5 y+54}-18}{5}-9$

$=\frac{5 y+63-18-45}{5}$

$=y$

$=I_{Y}$, Identity function

Also, $g o f(x)=g(f(x))$

$=g\left(5 x^{2}+6 x-9\right)$

$=\frac{\sqrt{5\left(5 x^{2}+6 x-9\right)+54-3}}{5}$

$=\frac{\sqrt{25 x^{2}+30 x-45+54-3}}{5}$

$=\frac{\sqrt{25 x^{2}+30 x+9-3}}{5}$

$=\frac{\sqrt{(5 x+3)^{2}-3}}{5}$

$=\frac{5 x+3-3}{5}$

$=\frac{5 x+3-3}{5}$

$=x$

$=I_{X}$, Identity function

So, $f$ is invertible.

Also, $f^{-1}(y)=g(y)=\frac{\sqrt{5 y+54}-3}{5}$