Aus Online Mathematik Brückenkurs 2

Version vom 12:59, 17. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 12:55, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
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
-	{{Displayed math||<math>f^{\,\prime}(x) = 6x^2+6x-12 = 0\,\textrm{.}</math>}}	+	{{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
-	{{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
-	{{Displayed math||<math>\Bigl(x+\frac{1}{2}\Bigr)^2 = \frac{9}{4}</math>}}	+	{{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
-	{{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{.}

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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 \displaystyle f^{\,\prime}(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

\displaystyle f^{\,\prime}(x) = 6x^2+6x-12 = 0\,\textrm{.}

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

\displaystyle \Bigl(x+\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 - 2 = 0\,\textrm{.}

This gives us the equation

\displaystyle \Bigl(x+\frac{1}{2}\Bigr)^2 = \frac{9}{4}

and taking the square root gives the solutions

\displaystyle \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} = 1\,\textrm{.} \end{align}

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

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

\displaystyle x		\displaystyle -2		\displaystyle 1
\displaystyle f^{\,\prime}(x)	\displaystyle +	\displaystyle 0	\displaystyle -	\displaystyle 0	\displaystyle +
\displaystyle f(x)	\displaystyle \nearrow	\displaystyle 21	\displaystyle \searrow	\displaystyle -6	\displaystyle \nearrow

The function has a local maximum at \displaystyle x=-2 and a local minimum at \displaystyle 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.