# 3. (3 points) Find the distance between the curves $y=e^{3 x+11}$ and $y=(\ln x-11) / 3$.
Solution: The distance between the graphs of the functions is equal to the distance between their nearest points. Note that the functions, between the graphs of which we need to find the distance, are inverses of each other, and their graphs are symmetric with respect to the line $y=x$, and, moreover, are on opposite sides of it.

If we find a point $A$ on the graph of the function $y=e^{3 x+11}$ that is closest to the line $y=x$, then the point $A^{\prime}$, symmetric to it, will obviously be the closest to this line on the graph of the function $y=(\ln x-11) / 3$. If we draw lines through these points parallel to the line $y=x$, the graphs of both functions will be outside the strip bounded by these lines, meaning that the distance between any two points on these graphs is not less than the distance between these lines. On the other hand, the distance between these lines is equal to the distance between the points $A$ and $A^{\prime}$, so $A A^{\prime}$ is the desired distance.

Let the coordinates of point $A$ be $\left(x ; e^{3 x+11}\right)$, then $A^{\prime}$ has coordinates $\left(e^{3 x+11} ; x\right)$ and the distance between them

$$
\rho(x)=\sqrt{\left(x-e^{3 x+11}\right)^{2}+\left(e^{3 x+11}-x\right)^{2}}=\sqrt{2\left(e^{3 x+11}-x\right)^{2}}=\sqrt{2}\left|e^{3 x+11}-x\right|
$$

Consider the function $f(x)=e^{3 x+11}-x$ and find its minimum using the derivative: $f^{\prime}(x)=3 e^{3 x+11}-1$. Setting this derivative to zero at point $x_{0}$, we get $x_{0}=$ $\frac{-\ln 3-11}{3}$.

Then for $xx_{0}$ - it increases, hence the function $f(x)$ reaches its minimum at point $x_{0}$.

$f\left(x_{0}\right)=\frac{1}{3}+\frac{\ln 3+11}{3}>0$, therefore $f(x)>0$ for $x \in R$.

Thus, $\rho(x)=\sqrt{2}\left(e^{3 x+11}-x\right)$.

Note that $\rho(x)$ differs from $f(x)$ by multiplication by $\sqrt{2}$, hence the minimum of the function is achieved at the same point as the minimum of the function $f(x)$. Therefore, the minimum distance is: $\sqrt{2}\left(4+\frac{\ln 3}{3}\right)$.