Find the intervals of monotonicity of y = ln x. What is
Find the intervals of monotonicity of $y = \ln x$. What is the maximum value of the function on $\bigl[2,\,e\bigr]$?
Explanation of Method
A monotone function needs no critical points.
Step 1: Domain
The logarithm is defined for $x\gt 0$, so the domain is $\bigl(0,+\infty\bigr)$ — a single interval, with no breaks.
Step 2: The derivative
\[ y' = \frac1x. \tag{1} \]
Step 3: Sign
On the domain $x\gt 0$, so $y' = \dfrac1x\gt 0$ everywhere. The derivative never vanishes: there are no critical points at all.
Step 4: Monotonicity
Hence $y = \ln x$ is strictly increasing on the whole of $\bigl(0,+\infty\bigr)$.
Step 5: Maximum on $\bigl[2,\,e\bigr]$
A strictly increasing function on a closed interval attains its greatest value at the right endpoint: \[y_{\max} = \ln e = 1, \qquad\text{and for comparison}\qquad y_{\min} = \ln2 = 0.6931472.\]
Step 6: Verification
\[\ln2 = 0.6931472 \lt \ln2.5 = 0.9162907 \lt \ln e = 1 \;\;\checkmark\]
\[y'' = -\frac{1}{x^{2}} \lt 0:\ \text{the growth slows, but never stops} \;\;\checkmark\]
\[\text{The maximum is at an endpoint, as it must be when } y'\neq0 \text{ anywhere} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\ln x \text{ is strictly increasing on } \bigl(0,+\infty\bigr); \qquad \max_{[2,\,e]}\ln x = \ln e = 1\,}\]
Remark
Remark. The point of the exercise is that the search for critical points may end empty-handed and the problem still be solved: on a closed interval a continuous function always attains its extreme values, and when the derivative has no zeros those values are at the ends. Note also that $\ln x\to-\infty$ as $x\to0^{+}$ and $\ln x\to+\infty$ as $x\to\infty$, so on the open domain neither a maximum nor a minimum exists — the closed interval is what makes the question meaningful.