Sketch f(x) = |ln x | and determine which condition

Sketch $f(x) = \bigl|\ln x\bigr|$ and determine which condition of Lagrange's theorem is not satisfied on $\left[\dfrac{1}{e},\,e\right]$.

Step 1: The function in pieces

Since $\ln x\lt 0$ on $\bigl(0,1\bigr)$ and $\ln x\ge0$ for $x\ge1$, \[f(x) = \begin{cases} -\ln x, & \tfrac1e \le x \lt 1,\\ \ \ \ln x, & 1\le x\le e. \end{cases}\]

Step 2: Continuity holds

Both branches are continuous, and they agree at the join: \[\lim_{x\to1^{-}}\bigl(-\ln x\bigr) = 0 = \lim_{x\to1^{+}}\ln x = f(1).\]

Step 3: Differentiability fails at $x = 1$

\[ f'_{-}(1) = \lim_{h\to0^{-}}\frac{-\ln\bigl(1+h\bigr)}{h} = -1, \qquad f'_{+}(1) = \lim_{h\to0^{+}}\frac{\ln\bigl(1+h\bigr)}{h} = +1. \]

The one-sided derivatives differ, so $f'(1)$ does not exist and $x = 1$ is an interior point of $\left(\tfrac1e,e\right)$.

Step 4: The endpoint values are equal — and the conclusion fails

It is worth computing the chord slope carefully, because the absolute value is easy to overlook: \[f\left(\frac1e\right) = \left|\ln\frac1e\right| = \bigl|-1\bigr| = 1, \qquad f(e) = \bigl|\ln e\bigr| = 1.\]

So $f\left(\tfrac1e\right) = f(e)$ and the chord is horizontal:

\[ \frac{f(e)-f\left(\tfrac1e\right)}{e-\tfrac1e} = \frac{1-1}{e-\tfrac1e} = 0. \]

The mean value theorem would therefore demand a point with $f'(\xi) = 0$. But away from $x = 1$, $f'(x) = \mp\frac1x \neq 0 \ \text{ for every } x$.

and at $x = 1$ the derivative does not exist. Hence no such $\xi$ exists: the conclusion genuinely fails, not merely the hypothesis.

Step 5: Verification

\[\frac1e \approx 0.3679:\ f = 1; \qquad x = 1:\ f = 0; \qquad e \approx 2.7183:\ f = 1 \;\;\checkmark\]

\[f'(x) = -\frac1x \in \bigl(-e,-1\bigr) \text{ on the left branch}, \qquad f'(x) = \frac1x \in \left(\frac1e,1\right) \text{ on the right}\]

\[\text{Neither range contains } 0 \;\;\checkmark\]