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Lösung 1.3:3d

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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<math>1</math>. The function is thus defined and differentiable everywhere. In order to determine the critical points, we differentiate the function using the quotient rule,
<math>1</math>. The function is thus defined and differentiable everywhere. In order to determine the critical points, we differentiate the function using the quotient rule,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
f^{\,\prime}(x)
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{\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]
Zeile 18: Zeile 18:
The derivative is zero when the numerator is zero and this gives us the equation
The derivative is zero when the numerator is zero and this gives us the equation
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{{Displayed math||<math>2x\bigl(1-2x^2-x^4\bigr) = 0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>2x\bigl(1-2x^2-x^4\bigr) = 0\,\textrm{.}</math>}}
The left-hand side is zero when one of the factors, <math>x</math> or <math>1-2x^2-x^4</math> is zero, i.e. either <math>x=0</math> or
The left-hand side is zero when one of the factors, <math>x</math> or <math>1-2x^2-x^4</math> is zero, i.e. either <math>x=0</math> or
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{{Displayed math||<math>1 - 2x^2 - x^4 = 0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>1 - 2x^2 - x^4 = 0\,\textrm{.}</math>}}
The last equation is a second-degree equation in <math>x^2</math>, which is perhaps simpler to see if we substitute <math>t=x^{2}</math>,
The last equation is a second-degree equation in <math>x^2</math>, which is perhaps simpler to see if we substitute <math>t=x^{2}</math>,
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{{Displayed math||<math>1-2t-t^{2}=0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>1-2t-t^{2}=0\,\textrm{.}</math>}}
The solutions are obtained by completing the square,
The solutions are obtained by completing the square,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
t^2 + 2t - 1 &= 0\,,\\[5pt]
t^2 + 2t - 1 &= 0\,,\\[5pt]
(t+1)^2 - 1^2 - 1 &= 0\,,\\[5pt]
(t+1)^2 - 1^2 - 1 &= 0\,,\\[5pt]
Zeile 44: Zeile 44:
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
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
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{{Displayed math||<math>f^{\,\prime}(x) = \frac{2x\bigl(1-2x^2-x^4\bigr)}{\bigl(1+x^4\bigr)^2}</math>}}
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{{Abgesetzte Formel||<math>f^{\,\prime}(x) = \frac{2x\bigl(1-2x^2-x^4\bigr)}{\bigl(1+x^4\bigr)^2}</math>}}
and by completing the square of the expression <math>1-2x^2-x^4</math> with respect to <math>x^{2}</math>,
and by completing the square of the expression <math>1-2x^2-x^4</math> with respect to <math>x^{2}</math>,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
1-2x^2-x^4 &= 1-\bigl(2x^2+x^4\bigr)\\[5pt]
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]
&= 1-\bigl(\bigl(x^2+1\bigr)^2-1^2\bigr)\\[5pt]
Zeile 56: Zeile 56:
we can write the derivative in the form
we can write the derivative in the form
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{{Displayed math||<math>f^{\,\prime}(x) = \frac{2x\bigl(2-\bigl(x^2+1\bigr)^2\bigr)}{\bigl(1+x^4\bigr)^2}</math>}}
+
{{Abgesetzte Formel||<math>f^{\,\prime}(x) = \frac{2x\bigl(2-\bigl(x^2+1\bigr)^2\bigr)}{\bigl(1+x^4\bigr)^2}</math>}}
where it is rather simple to determine the sign of the individual factors.
where it is rather simple to determine the sign of the individual factors.

Version vom 12:56, 10. Mär. 2009

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

  1. critical points, i.e. where f(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 1+x4 is, however, a sum of the number 1 and x4 which is always positive (x4 is the square of x2), and hence the denominator is always greater than or equal to 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 
\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.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:3d“