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Aus Online Mathematik Brückenkurs 2

Lokale Extrempunkte einer Funktion sind entweder:

1. stationäre Punkte, wo f(x)=0,
2. Singuläre Punkte, wo die Funktion nicht ableitbar ist, oder
3. Endpunkte.

Wir untersuchen zuerst die Bedienungen 2 und 3. Die Funktion besteht aus einen Bruch von zwei Polynomen. Die Funktion ist nur undefiniert wenn der Nenner null ist. Nachdem der Nenner 1+x4 ist, wird er immer positiv. 1. The function is thus defined and differentiable everywhere. In order to determine the critical points, we differentiate the function using the quotient rule,

f(x)=1+x421+x21+x4−1+x21+x4=1+x422x1+x4−1+x24x3=1+x422x+2x5−4x3−4x5=1+x422x1−2x2−x4.

The derivative is zero when the numerator is zero and this gives us the equation

2x1−2x2−x4=0.

The left-hand side is zero when one of the factors, x or 1−2x2−x4 is zero, i.e. either x=0 or

1−2x2−x4=0.

The last equation is a second-degree equation in x2, which is perhaps simpler to see if we substitute t=x2,

1−2t−t2=0.

The solutions are obtained by completing the square,

t2+2t−1(t+1)2−12−1(t+1)2=0=0=2

and are t=−12 . It is only one of these solutions, t=−1+2 ,that is positive and can be equal to x2.

The function has therefore three critical points, x=−2−1 , x=0 and x=2−1 .

We can determine the character of the critical points by writing down the sign of its derivative. It is useful to write down the derivative in an appropriately factorized form first. We know already that

f(x)=1+x422x1−2x2−x4

and by completing the square of the expression 1−2x2−x4 with respect to x2,

1−2x2−x4=1−2x2+x4=1−x2+12−12=2−x2+12

we can write the derivative in the form

f(x)=1+x422x2−x2+12

where it is rather simple to determine the sign of the individual factors.

x		−2−1		0		2−1
2x	−	−	−	\displaystyle 0	\displaystyle +	\displaystyle +	\displaystyle +
\displaystyle 2 - (x^2 + 1)^2	\displaystyle -	\displaystyle 0	\displaystyle +	\displaystyle +	\displaystyle +	\displaystyle 0	\displaystyle -
\displaystyle (x^4 + 1)^2	\displaystyle +	\displaystyle +	\displaystyle +	\displaystyle +	\displaystyle +	\displaystyle +	\displaystyle +

If we multiply these factors together, we get an outline of the derivative's sign and can draw conclusions about whether the critical points are local maximum points, minimum points or neither.

\displaystyle x		\displaystyle -\sqrt{ \sqrt{2} - 1}		\displaystyle 0		\displaystyle \sqrt{ \sqrt{2} - 1}
\displaystyle \insteadof{2 - (x^2 + 1)^2}{f^{\, \prime} (x)}	\displaystyle +	\displaystyle 0	\displaystyle -	\displaystyle 0	\displaystyle +	\displaystyle 0	\displaystyle -
\displaystyle f(x)	\displaystyle \nearrow	\displaystyle \tfrac{1 }{2} (\sqrt{2} + 1)	\displaystyle \searrow	\displaystyle 1	\displaystyle \nearrow	\displaystyle \tfrac{1 }{2} (\sqrt{2} + 1)	\displaystyle \searrow

The function has local maximum points at \displaystyle x=\pm \sqrt{\sqrt{2}-1} and a local minimum at \displaystyle x=0.