Corner point of |ln x|

Problem

Verify that $f(x) = |\ln x|$ has a corner point at $x = 1$ and compute the two half-tangents.

Solution

At $x=1$ we have $\ln 1 = 0$, hence $f(1)=0$ and $f$ is continuous. I distinguish the two sides:

• For $x>1$: $\ln x>0$, so $f=\ln x$, $f'(x)=1/x$, whence $f'_+(1)=1$.
• For $x<1$ (with $x>0$): $\ln x<0$, so $f=-\ln x$, $f'(x)=-1/x$, whence $f'_-(1)=-1$.

The one-sided derivatives are $1$ and $-1$: finite and different, therefore a corner point. The two half-tangents at $(1,0)$ are: $y = 1\cdot(x-1) = x-1 \quad (\text{right}), \qquad y = -1\cdot(x-1) = 1-x \quad (\text{left}).$

Links

Topics: Derivatives
Concepts: One-sided derivative · Corner point
Methods: Non-differentiable corner point
Skills: Reason by cases
Exercise type: Derivative computation