x^{-\frac{1}{6}}\leq x^{-\frac{1}{8}}
\frac{1}{x^{\frac{1}{6}}}-\frac{1}{x^{\frac{1}{8}}}\leq 0 , założenie x\ne 0
\frac{1}{x^{\frac{4}{24}}}-\frac{1}{x^{\frac{3}{24}}}\leq 0
\frac{1-x^{\frac{1}{24}}}{x^{\frac{4}{24}}}\leq 0
\frac{1-\sqrt[24]{x}}{x^{\frac{1}{6}}}\leq 0
-\frac{\sqrt[24]{x}-1}{\sqrt[6]{x}}\leq 0 \ |*(-1)
\frac{\sqrt[24]{x}-1}{\sqrt[6]{x}}\geq 0
Wyznaczam miejsca zerowe
f(x)=0
\frac{\sqrt[24]{x}-1}{\sqrt[6]{x}}=0
\sqrt[24]{x}-1=0 \Rightarrow x=1
\sqrt[24]{x}-1\geq 0
x\in \langle1;\infty)
sprawdzenie
\frac{1-1}{1}\geq 0