f:R₊ [9,infinity]
f(x)=5x²+6x-9.
Prove that f is invertible
USING COMPLETING SQUARE METHOD ONLY

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{f}:{\mathrm{R}}_{+}\to \left[-9,\infty \right] \left(\mathrm{it} \mathrm{should} \mathrm{be} -9\right) \\ \left[\mathbf{-}\mathbf{9}\mathbf{,}\mathbf{\infty }\right]\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{called}\mathbf{ }\mathbf{co}\mathbf{-}\mathbf{domain}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{f}\left(\mathbf{x}\right) \\ \mathrm{f}\left(\mathrm{x}\right)=5{\mathrm{x}}^2+6\mathrm{x}-9 \\ \mathrm{since} \mathrm{domain} \mathrm{is} {\mathrm{R}}_{+}, \mathrm{x}\ge 0 \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=10\mathrm{x}+6 \mathrm{since} \mathrm{x}>0, \mathrm{f}\text{'}\left(\mathrm{x}\right)>0, \mathrm{hence} \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{increasing} \mathrm{or} \mathrm{one}-\mathrm{one}. \\ \mathrm{now} \mathrm{range} \mathrm{of} \mathrm{function} \mathrm{will} \mathrm{be}: \\ \mathrm{since} \mathrm{function} \mathrm{is} \mathrm{increasing} \mathrm{in} \left[0,\infty \right], \mathrm{we} \mathrm{can} \mathrm{say} \\ \mathrm{lowest} \mathrm{value} \mathrm{will} \mathrm{be} \mathrm{at} \mathrm{x}=0 \\ \mathrm{hence}, \\ \mathrm{f}\left(\mathrm{x}\right)=0+0-9=-9 \\ \mathrm{hence} \mathrm{range} \mathrm{for} {\mathrm{R}}_{+} \mathrm{is} \left[-9,\infty \right] \\ \mathrm{since} \mathrm{range} = \mathrm{co}-\mathrm{domain} \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{onto}. \\ \mathrm{Hence} \mathrm{it} \mathrm{is} \mathrm{invertible}: \\ \mathrm{now} \mathrm{by} \mathbf{completing}\mathbf{ }\mathbf{square}\mathbf{ }\mathbf{method}\mathbf{:} \\ \mathrm{f}\left(\mathrm{x}\right)=5{\mathrm{x}}^2+6\mathrm{x}-9=5\left({\mathrm{x}}^2+\frac{6\mathrm{x}}{5}-\frac{9}{5}\right)=5\left({\mathrm{x}}^2+2.\mathrm{x}.\frac{3}{5}+\frac{9}{25}-\frac{9}{5}-\frac{9}{25}\right) \left[\mathrm{adding} \mathrm{and} \mathrm{subtracting} \frac{9}{25}\right] \\ \mathrm{y}=5\left({\mathrm{x}}^2+2.\mathrm{x}.\frac{3}{5}+{\left(\frac{3}{5}\right)}^2-\frac{54}{25}\right) \\ \mathrm{y}=5{\left(\mathrm{x}+\frac{3}{5}\right)}^2-\frac{54}{5} \\ \mathrm{hence}, \\ \mathrm{y}+\frac{54}{5}=5{\left(\mathrm{x}+\frac{3}{5}\right)}^2 \\ \frac{\mathrm{y}}{5}+\frac{54}{25}={\left(\mathrm{x}+\frac{3}{5}\right)}^2 \\ \mathrm{x}+\frac{3}{5}=\pm \sqrt{\frac{\mathrm{y}}{5}+\frac{54}{25}} \\ \mathrm{x}=-\frac{3}{5}\pm \sqrt{\frac{\mathrm{y}}{5}+\frac{54}{25}} \\ \mathrm{since} \mathrm{x}>0 \\ \mathrm{we} \mathrm{will} \mathrm{take} \mathrm{only}: \\ \mathrm{x}=\frac{-3}{5}+\sqrt{\frac{\mathrm{y}}{5}+\frac{54}{25}} \\ \mathbf{5}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{+}\sqrt{\mathbf{5}\mathbf{y}\mathbf{+}\mathbf{54}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ans}\mathbf{ } \end{gathered}$$

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