The minimum value of

Since

Arithmetic mean ≥ geometric mean of 4x and 41 – x

i. e $\frac{4^{x}+4^{-x}}{2} \geq \sqrt{4^{x}} \cdot 4^{1-x}$

i. e $\frac{4^{x}+4^{1-x}}{2} \geq \sqrt{4^{x} \cdot 4 \cdot 4^{-x}}$

i. e $\frac{4^{x}+4^{1-x}}{2} \geq 2$

i. e $4^{x}+4^{1-x} \geq 2 \times 2=4$

i.e minimum value of 4x + 41 – x is 4.