Problem 7. Solve the inequality $\sqrt[4]{x-1}+$ $+\sqrt[4]{x+14} \geqslant 3$
Solution. The left side of the inequality is defined on the interval $[1 ;+\infty)$. The function $f_{1}(x)=\sqrt[4]{x-1}$ is continuous and monotonic (on the specified interval it increases from 0 to $+\infty$). The function $f_{2}(x)=\sqrt[4]{x+14}$ is continuous and monotonic (on the interval $[1 ;+\infty$) it increases from $\sqrt[4]{15}$ to $+\infty$), therefore the continuous function $f(x)=f_{1}(x)+f_{2}(x)$ on the half-interval $[1 ;+\infty)$ increases from $\sqrt[4]{15}$ to $+\infty$. Since $\sqrt[4]{15}<3$, there exists a unique point $x_{0}$ such that $f\left(x_{0}\right)=3$. Obviously, $x_{0}=2$. Answer. $x \geqslant 2$.