6. Find the set of values of the function $f(x)=1 / g(64 g(g(\ln x)) / 1025)$, where

$$
g(x)=x^{5}+1 / x^{5} \cdot(10 \text { points })
$$
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 the sign 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 $t=1$, $t=1$ 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

![](https://cdn.mathpix.com/cropped/2024_05_06_50f0bbe00f908a027b9ag-04.jpg?height=885&width=805&top_left_y=1057&top_left_x=1162)
is the set $(-\infty ;-2] \cup[2 ;+\infty)$. The function $g(x)=x^{5}+1 / x^{5}=\varphi\left(x^{5}\right)$. Since the function $t=x^{5}$ is increasing on the interval $(-\infty ;+\infty)$ and takes all numerical values, the set of values of the function $\varphi\left(x^{5}\right)$, 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 $g(\ln x)$ is also the set $(-\infty ;-2] \cup[2 ;+\infty)$. The set of values of the function $g(g(\ln x))$ is the set $(-\infty ;-1025 / 32] \cup[1025 / 32 ;+\infty)$, and the function $64 g(g(\ln x)) / 1025-$ has the set of values $(-\infty ;-2] \cup[2 ;+\infty)$.

Thus, $g(64 g(g(\ln x)) / 1025) \in(-\infty ;-1025 / 32] \cup[1025 / 32 ;+\infty)$. From this, we find the set $E_{f}$ of values of the function $f(x)=1 / g(64 g(g(\ln x)) / 1025): E_{f}=[-32 / 1025 ; 0) \cup(0 ; 32 / 1025]$.

Answer: $E_{f}=[-32 / 1025 ; 0) \cup(0 ; 32 / 1025]$.