Lösung 1.3:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 1.3:2c moved to Solution 1.3:2c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
There are three types of points at which the function can have local extreme points:
-
<center> [[Image:1_3_2c-1(3).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
1. Critical points, i.e. where
-
{{NAVCONTENT_START}}
+
<math>{f}'\left( x \right)=0</math>;
-
<center> [[Image:1_3_2c-2(3).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
2. Points where the function is not differentiable;
-
{{NAVCONTENT_START}}
+
 
-
<center> [[Image:1_3_2c-3(3).gif]] </center>
+
3. Endpoints of the interval of definition.
-
{{NAVCONTENT_STOP}}
+
 
 +
Because our function is a polynomial, it is defined and differentiable everywhere, and therefore does not have any points which satisfy
 +
<math>2</math>
 +
and
 +
<math>3</math>.
 +
 
 +
As regards
 +
<math>1</math>, we set the derivative equal to zero and obtain the equation
 +
 
 +
 
 +
<math>{f}'\left( x \right)=6x^{2}+6x-12=0</math>
 +
 
 +
Dividing both sides by
 +
<math>6</math>
 +
and completing the square, we obtain
 +
 
 +
 
 +
<math>\left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}-2=0</math>
 +
 
 +
This gives us the equation
 +
 
 +
 
 +
<math>\left( x+\frac{1}{2} \right)^{2}=\frac{9}{4}</math>
 +
 
 +
 
 +
and taking the root gives the solutions
 +
 
 +
 
 +
<math>x=-\frac{1}{2}-\sqrt{\frac{9}{4}}=-\frac{1}{2}-\frac{3}{2}=-2</math>
 +
 
 +
 
 +
<math>x=-\frac{1}{2}+\sqrt{\frac{9}{4}}=-\frac{1}{2}+\frac{3}{2}=1</math>
 +
 
 +
 
 +
This means that if the function has several extreme points, they must lie between
 +
<math>x=-2\text{ }</math>
 +
and
 +
<math>x=1</math>.
 +
 
 +
Then, we write down a sign table for the derivative, and read off the possible extreme points.
 +
 
 +
TABLE
 +
 
 +
The function has a local maximum at
 +
<math>x=-2\text{ }</math>
 +
and a local minimum at
 +
<math>x=1</math>.
 +
 
 +
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.
 +
 
 +
PICTURE TABLE
 +
 
 +
 
[[Image:1_3_2_c.gif|center]]
[[Image:1_3_2_c.gif|center]]

Version vom 11:01, 15. Okt. 2008

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

1. Critical points, i.e. where \displaystyle {f}'\left( x \right)=0;

2. Points where the function is not differentiable;

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 \displaystyle 2 and \displaystyle 3.

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


\displaystyle {f}'\left( x \right)=6x^{2}+6x-12=0

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


\displaystyle \left( x+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}-2=0

This gives us the equation


\displaystyle \left( x+\frac{1}{2} \right)^{2}=\frac{9}{4}


and taking the root gives the solutions


\displaystyle x=-\frac{1}{2}-\sqrt{\frac{9}{4}}=-\frac{1}{2}-\frac{3}{2}=-2


\displaystyle x=-\frac{1}{2}+\sqrt{\frac{9}{4}}=-\frac{1}{2}+\frac{3}{2}=1


This means that if the function has several extreme points, they must lie between \displaystyle x=-2\text{ } and \displaystyle x=1.

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

TABLE

The function has a local maximum at \displaystyle x=-2\text{ } 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.

PICTURE TABLE


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