6. Find the set of values of the function $f(x)=1 / g\left(64 g\left(16 g\left(\log _{2} x\right)\right) / 5\right)$, where

$g(x)=\sqrt[5]{x}+1 / \sqrt[5]{x}$.
Solution: First, consider the function $\varphi(t)=t+1 / t$. The function $\varphi(t)$ is defined for all $t \neq 0$. Let's find the extrema of the function $\varphi(t)$. To do this, we will determine the intervals of constancy of the derivative of the function $\varphi(t): \quad \varphi^{\prime}(t)=1-1 / t^{2}=(t-1)(t+1) / t^{2}$, $\varphi^{\prime}(t)=0$ when $t= \pm 1$. As the derivative $\varphi^{\prime}(t)$ changes sign from positive to negative when passing through the point $t=-1$, $t=-1$ is a point of maximum, $\varphi_{\max }=\varphi(-1)=-2$. As the derivative $\varphi^{\prime}(t)$ changes sign from negative to positive when passing through the point $t=1$

![](https://cdn.mathpix.com/cropped/2024_05_06_50f0bbe00f908a027b9ag-12.jpg?height=885&width=788&top_left_y=203&top_left_x=1134)
$t=1$ is a point of minimum, $\varphi_{\min }=\varphi(1)=2$. The graph of the function $\varphi(t)$ is shown in the figure. The set of values of this function is the set $(-\infty ;-2] \cup[2 ;+\infty)$. The function $g(x)=\sqrt[5]{x}+1 / \sqrt[5]{x}=\varphi(\sqrt[5]{x})$. Since the function $t=\sqrt[5]{x}$ is increasing on the interval $(-\infty ;+\infty)$ and takes all numerical values, the set of values of the function $\varphi(\sqrt[5]{x})$, and therefore of $g(x)$, is the set $(-\infty ;-2] \cup[2 ;+\infty)$, with $g_{\max }=g(-1)=-2, \quad g_{\min }=g(1)=2$. For the same reason, the set of values of the function $g\left(\log _{2} x\right)$ is also the set $(-\infty ;-2] \cup[2 ;+\infty)$. The set of values of the function $g\left(16 g\left(\log _{2} x\right)\right)$ is the set $(-\infty ;-5 / 2] \cup[5 / 2 ;+\infty)$, and the function $64 g\left(16 g\left(\log _{2} x\right)\right) / 5-$ is the set $(-\infty ;-32] \cup[32 ;+\infty)$. Thus, $g\left(64 g\left(16 g\left(\log _{2} x\right)\right) / 5\right) \in(-\infty ;-5 / 2] \cup[5 / 2 ;+\infty)$. From this, we find the set $E_{f}$ of values of the function $f(x)=1 / g\left(64 g\left(16 g\left(\log _{2} x\right)\right) / 5\right): E_{f}=[-2 / 5 ; 0) \cup(0 ; 2 / 5]$.

Answer: $E_{f}=[-2 / 5 ; 0) \cup(0 ; 2 / 5]$.