Question

The minimum value of 4^x + 4^{1 – x}, x ∈ R, is ___________.

Answer 1

Since
Arithmetic mean ≥ geometric mean of 4^x and 4^{1 – x}
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{4^x+4^{1-x}}{2}\ge \sqrt{4^x.4^{1-x}} \\ \mathrm{i}.\mathrm{e}\frac{4^x+4^{1-x}}{2}\ge \sqrt{4^x.4.4^{-x}} \\ \mathrm{i}.\mathrm{e}\frac{4^x+4^{1-x}}{2}\ge 2 \\ \mathrm{i}.\mathrm{e}{4}^x+4^{1-x}\ge 2\times 2=4 \end{gathered}$$

i.e minimum value of 4^x + 4^{1 – x} is 4.