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Lösung 1.3:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 9: Zeile 9:
As regards item 1, we set the derivative equal to zero and obtain the equation
As regards item 1, we set the derivative equal to zero and obtain the equation
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{{Displayed math||<math>f^{\,\prime}(x) = 6x^2+6x-12 = 0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>f^{\,\prime}(x) = 6x^2+6x-12 = 0\,\textrm{.}</math>}}
Dividing both sides by 6 and completing the square, we obtain
Dividing both sides by 6 and completing the square, we obtain
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{{Displayed math||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 = 0\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 = 0\,\textrm{.}</math>}}
This gives us the equation
This gives us the equation
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{{Displayed math||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 = \frac{9}{4}</math>}}
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{{Abgesetzte Formel||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 = \frac{9}{4}</math>}}
and taking the square root gives the solutions
and taking the square root gives the solutions
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x &= -\frac{1}{2}-\sqrt{\frac{9}{4}} = -\frac{1}{2}-\frac{3}{2} = -2\,,\\[5pt]
x &= -\frac{1}{2}-\sqrt{\frac{9}{4}} = -\frac{1}{2}-\frac{3}{2} = -2\,,\\[5pt]
x &= -\frac{1}{2}+\sqrt{\frac{9}{4}} = -\frac{1}{2}+\frac{3}{2} = 1\,\textrm{.}
x &= -\frac{1}{2}+\sqrt{\frac{9}{4}} = -\frac{1}{2}+\frac{3}{2} = 1\,\textrm{.}

Version vom 12:55, 10. Mär. 2009

There are three types of points at which the function can have local extreme points,

  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.

Because our function is a polynomial, it is defined and differentiable everywhere, and therefore does not have any points which satisfy items 2 and 3.

As regards item 1, we set the derivative equal to zero and obtain the equation

f(x)=6x2+6x−12=0.

Dividing both sides by 6 and completing the square, we obtain

x+212−212−2=0. 

This gives us the equation

x+212=49 

and taking the square root gives the solutions

xx=−21−49=−21−23=−2=−21+49=−21+23=1.

This means that if the function has several extreme points, they must be among x=−2 and x=1.

Then, we write down a sign table for the derivative, and read off the possible extreme points.

x −2 1
f(x) + 0 − 0 +
f(x) 21 −6


The function has a local maximum at x=−2 and a local minimum at x=1.

We obtain the overall appearance of the graph of the function from the table and by calculating the value of the function at a few points.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.3:2c“