Aus Online Mathematik Brückenkurs 2

The only points which can possibly be local extreme points of the function are one of the following:

1. Critical points, i.e. where \displaystyle {f}'\left( x \right)=0;

2. Points where the function is not differentiable;

3. Endpoints of the interval of definition.

What determines the function's region of definition is \displaystyle \ln x, which is defined for \displaystyle x>0, and this region does not have any endpoints ( \displaystyle x=0 does not satisfy \displaystyle x>0 ), so item 3. above does not give rise to any imaginable extreme points. Furthermore, the function is differentiable everywhere (where it is defined), because it consists of \displaystyle x and \displaystyle \ln x which are differentiable functions; so, item 2. above does not contribute any extreme points either.

All the remains are possibly critical points. We differentiate the function

\displaystyle {f}'\left( x \right)=1\centerdot \ln x+x\centerdot \frac{1}{x}-0=\ln x+1

and see that the derivative is zero when

\displaystyle \ln x=-1\quad \Leftrightarrow \quad x=e^{-1}

In order to determine whether this is a local maximum, minimum or saddle point, we calculate the second derivative, \displaystyle {f}''\left( x \right)={1}/{x}\; which gives that

\displaystyle {f}''\left( e^{-1} \right)=\frac{1}{e^{-1}}=e>0

which implies that \displaystyle x=e^{-1} is a local minimum.