If x  be real, then the maximum value of $5+4\mathrm{x}-4{\mathrm{x}}^2$  will be equal to

Let $\mathrm{f}\left(\mathrm{x}\right)=5+4\mathrm{x}-4{\mathrm{x}}^2=\mathrm{y}\Rightarrow 4{\mathrm{x}}^2-4\mathrm{x}-5+\mathrm{y}=0$

Since x is real, so ${\mathrm{B}}^2-4\mathrm{AC}\ge 0$

$\begin{array}{rl} & \Rightarrow 16-4.4(-5+\mathrm{y})\ge 0\\ & \Rightarrow 16-16(-5+\mathrm{y})\ge 0\Rightarrow -5+\mathrm{y}\le 1\Rightarrow \mathrm{y}\le 6\end{array}$

Hence maximum value of y is 6.