Aus Online Mathematik Brückenkurs 2

The local extreme points of the function are one of the following,

1. critical points, i.e. where \displaystyle f^{\,\prime}(x)=0,
2. points where the function is not differentiable, and
3. endpoints of the interval of definition.

We start with items 2 and 3. The function consists of a quotient of two polynomials, so the only possibility that the function is not defined or differentiable is if the denominator is zero somewhere. The denominator \displaystyle 1+x^{4} is, however, a sum of the number \displaystyle 1 and \displaystyle x^{4} which is always positive (\displaystyle x^{4} is the square of \displaystyle x^{2}), and hence the denominator is always greater than or equal to \displaystyle 1. The function is thus defined and differentiable everywhere. In order to determine the critical points, we differentiate the function using the quotient rule,

\displaystyle \begin{align} f^{\,\prime}(x) &= \frac{\bigl(1+x^2\bigr)^{\prime}\cdot\bigl(1+x^4\bigr) - \bigl(1+x^2\bigr)\cdot \bigl(1+x^4\bigr)^{\prime}}{\bigl(1+x^4\bigr)^2}\\[5pt] &= \frac{2x\bigl(1+x^4\bigr) - \bigl(1+x^2\bigr)4x^3}{\bigl(1+x^4\bigr)^2}\\[5pt] &= \frac{2x+2x^5-4x^3-4x^5}{\bigl(1+x^4\bigr)^2}\\[5pt] &= \frac{2x\bigl(1-2x^2-x^4\bigr)}{\bigl(1+x^4\bigr)^2}\,\textrm{.} \end{align}

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

\displaystyle 2x\bigl(1-2x^2-x^4\bigr) = 0\,\textrm{.}

The left-hand side is zero when one of the factors, \displaystyle x or \displaystyle 1-2x^2-x^4 is zero, i.e. either \displaystyle x=0 or

\displaystyle 1 - 2x^2 - x^4 = 0\,\textrm{.}

The last equation is a second-degree equation in \displaystyle x^2, which is perhaps simpler to see if we substitute \displaystyle t=x^{2},

\displaystyle 1-2t-t^{2}=0\,\textrm{.}

The solutions are obtained by completing the square,

\displaystyle \begin{align} t^2 + 2t - 1 &= 0\,,\\[5pt] (t+1)^2 - 1^2 - 1 &= 0\,,\\[5pt] (t+1)^2 &= 2\,, \end{align}

and are \displaystyle t=-1\pm \sqrt{2}. It is only one of these solutions, \displaystyle t=-1+\sqrt{2},that is positive and can be equal to \displaystyle x^2\,.

The function has therefore three critical points, \displaystyle x=-\sqrt{\sqrt{2}-1}, \displaystyle x=0 and \displaystyle x=\sqrt{\sqrt{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

\displaystyle f^{\,\prime}(x) = \frac{2x\bigl(1-2x^2-x^4\bigr)}{\bigl(1+x^4\bigr)^2}

and by completing the square of the expression \displaystyle 1-2x^2-x^4 with respect to \displaystyle x^{2},

\displaystyle \begin{align} 1-2x^2-x^4 &= 1-\bigl(2x^2+x^4\bigr)\\[5pt] &= 1-\bigl(\bigl(x^2+1\bigr)^2-1^2\bigr)\\[5pt] &= 2-\bigl(x^2+1\bigr)^2 \end{align}

we can write the derivative in the form

\displaystyle f^{\,\prime}(x) = \frac{2x\bigl(2-\bigl(x^2+1\bigr)^2\bigr)}{\bigl(1+x^4\bigr)^2}

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

\displaystyle x		\displaystyle -\sqrt{ \sqrt{2} - 1}		\displaystyle 0		\displaystyle \sqrt{ \sqrt{2} - 1}
\displaystyle 2x	\displaystyle -	\displaystyle -	\displaystyle -	\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.