Prove that a function $\displaystyle \mathrm{f}:[0, \infty) \rightarrow[-5, \infty)$ defined as $\displaystyle \mathrm{f}(\mathrm{x})=4 \mathrm{x}^{2}+4 \mathrm{x}-5$ is both one-one and onto.

Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in[0, \infty)$ such that $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)$ Then this $\displaystyle \Rightarrow 4 \mathrm{x}_{1}^{2}+4 \mathrm{x}_{1}-5=4 \mathrm{x}_{2}^{2}+4 \mathrm{x}_{2}-5$\Rightarrow\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)+\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)=0$\displaystyle \Rightarrow\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)\left[\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)+1\right]=0$ $\displaystyle \Rightarrow\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)=0$ or $\displaystyle \mathrm{x}_{1}+\mathrm{x}_{2}=-1$, which is rejected as $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \geq 0$ $\displaystyle \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$ $\displaystyle \therefore \mathrm{f}$ is one-one. Let $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{y} \Rightarrow \mathrm{y}=4 \mathrm{x}^{2}+4 \mathrm{x}-5$ for $\displaystyle \mathrm{x} \in[0, \infty)$\begin{aligned} & \Rightarrow 4 \mathrm{x}^{2}+4 \mathrm{x}-5-\mathrm{y}=0 & \Rightarrow \mathrm{x}=\frac{-4 \pm \sqrt{16-16(-5-\mathrm{y})}}{8} \Rightarrow \mathrm{x}=\frac{-4+4 \sqrt{6+\mathrm{y}}}{8}=\frac{-1+\sqrt{6+\mathrm{y}}}{2} \end{aligned}Since, $\displaystyle \mathrm{x} \geq 0$, we have $\displaystyle \mathrm{y}+6 \geq 1 \Rightarrow \mathrm{y} \in[-5, \infty)$ ∴ Range $\displaystyle =$ Codomain $\displaystyle =[-5, \infty)$ Hence f is onto.