Question

if f:R->R is defined by f(x)=a^x/a^x+roota, a>0, then evaluate f(x)+ f(1-x)

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(1-\mathrm{x}\right) \\ =\frac{{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{x}}+\sqrt{\mathrm{a}}}+\frac{{\mathrm{a}}^{1-\mathrm{x}}}{{\mathrm{a}}^{1-\mathrm{x}}+\sqrt{\mathrm{a}}}=\frac{{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{x}}+\sqrt{\mathrm{a}}}+\frac{{\displaystyle \frac{\mathrm{a}}{{\mathrm{a}}^{\mathrm{x}}}}}{\frac{\mathrm{a}}{{\mathrm{a}}^{\mathrm{x}}}+\sqrt{\mathrm{a}}} \\ =\frac{{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{x}}+\sqrt{\mathrm{a}}}+\frac{{\displaystyle \mathrm{a}}}{\mathrm{a}+{\mathrm{a}}^{\mathrm{x}}\sqrt{\mathrm{a}}}=\frac{{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{x}}+\sqrt{\mathrm{a}}}+\frac{{\displaystyle \sqrt{\mathrm{a}}\times \sqrt{\mathrm{a}}}}{\sqrt{\mathrm{a}}\times \sqrt{\mathrm{a}}+{\mathrm{a}}^{\mathrm{x}}\sqrt{\mathrm{a}}} \\ =\frac{{\mathrm{a}}^{\mathrm{x}}}{{\mathrm{a}}^{\mathrm{x}}+\sqrt{\mathrm{a}}}+\frac{{\displaystyle \sqrt{\mathrm{a}}}}{\sqrt{\mathrm{a}}+{\mathrm{a}}^{\mathrm{x}}}=\frac{{\mathrm{a}}^{\mathrm{x}}+\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}+{\mathrm{a}}^{\mathrm{x}}}=1 \end{gathered}$$